THE UNDENIABLE CHARM OF INEQUALITIES
Paul Vaderlind, Stockholm University
November, 2004
A Short Guided Tour through the Jungle of Algebraic Inequalities
Introduction ........................................................................1
Inequalities ........................................................................ 2
Problems 1........................................................................ 18
Proofs of the Inequalities ................................................. 23
Solutions to the Problems 1.............................................. 34
Problems 2 ....................................................................... 46
One more useful inequality .............................................. 49
Recent problems ............................................................... 51
INTRODUCTION
What follows is a list of eleven most useful algebraic inequalities with a large collection of examples and related problems. The inequalities are presented in a simple form: they may all be
strengthened and generalized. The choice of formulation is made in accordance with usefulness
for solving problems in mathematical competitions. Moreover, the inequalities presented here are
in no way “independent”. For example, Hölder’s inequality is a consequence of Jensen’s; both
Cauchy-Schwarz and Chebyshev’s inequalities may be derived from the Rearrangement inequality, and so on. Nevertheless, because of their applicability all these inequalities deserve to be stated
separately.
The inequalities presented are the following
1. AM-GM-HM inequality
2. Chebyshev’s inequality
3. Rearrangement inequality
4. Cauchy-Schwarz inequality
1
5. Hölder’s inequality
6. Minkowski’s inequality
7. Jensen’s inequality
8. Power Mean inequality
9. Schur’s inequality
10. Maclaurin’s inequality
11. Muirhead’s inequality
INEQUALITIES
1. AM-GM-HM inequality
Let a1 , a2 , ..., an be positive real numbers. Then
√
a1 + a2 + · · · + an
n
≥ n a1 a2 · · · an ≥
,
n
1/a1 + 1/a2 + · · · + 1/an
with equality if and only if a1 = a2 = · · · = an .
This probably is the best-known inequality and is very useful in various situations. At the same
time it is only a special case of several other inequalities listed below.
Example 1. (UK, 2000) Given that x, y, z are positive real numbers satisfying xyz = 32, find
the minimum value of x2 + 4xy + 4y 2 + 2z 2 .
2
2
Solution. Applying
+ 4xy + 4y 2 + 2z
= (x2 +
pthe AM-GM inequality twice, we find that x p
√
3
4y 2 ) + 4xy + 2z 2 ≥ 2 x2 · 4y 2 + 4xy + 2z 2 = 4xy + 4xy + 2z 2 ≥ 3 3 32x2 y 2 z 2 = 3 323 = 96.
Equality holds when x2 = 4y 2 and 4xy = 2z 2 , it is, when x = z = 4 and y = 2.
Example 2. (IMO, 1964) Let a, b, c be sides of a triangle. Prove that a2 (b + c − a) + b2 (c +
a − b) + c2 (a + b − c) ≤ 3abc.
Solution. Let x = a + b − c, y = b + c − a, z = c + a − b. Then x, y, z > 0 and the inequality
z + x 2
x + y 2
y + z 2
3
wa have to prove becomes
y+
z+
x ≤ (z + x)(x + y)(y + z).
2
2
2
8
This inequality reduces to x2 y + xy 2p+ x2 z + xz 2 + y 2 z + yz 2 ≥ 6xyz, which is true because
x2 y + xy 2 + x2 z + xz 2 + y 2 z + yz 2 ≥ 6 6 x6 y 6 z 6 = 6xyz by the AM-GM inequality. The equality
occurs if and only if x = y = z, i.e. if and only if the triangle is equilateral.
Example 3. Given that the equation x4 + px3 + qx2 + rx + s = 0 has four real positive roots,
prove that
(i)
pr − 16s ≥ 0,
2
(ii)
q 2 − 36s ≥ 0.
Solution. Suppose x1 , x2 , x3 and x4 are the roots of the equation. Then, we have x1 + x2 + x3 +
x4 = −p, x1 x2 +x1 x3 +x1 x4 +x2 x3 +x2 x4 +x3 x4 = q, x1 x2 x3 +x1 x2 x4 +x1 x3 x4 +x2 x3 x4 = −r
and x1 x2 x3 x4 = s.
4
4
X
X
1
x1 x2 x3 x4 ≥ 16s. From the AMUsing the AM-HM inequality we get pr =
xi
x
i
i=1
i=1
GM inequality we have q = x1 x2 + x1 x3 + x1 x4 + x2 x3 + x2 x4 + x3 x4 ≥ 6(x31 x32 x33 x34 )1/6 = 6s1/2 .
Example 4. (IMO, 1999) Let n ≥ 2 be a fixed integer. Find the smallest constant C such that
for all non-negative reals x1 , ..., xn :
X
xi xj (x2i
+
x2j )
≤C
n
X
xi
4
.
i=1
1≤i<j≤n
Determine when equality occurs.
Solution. We give here a surprise-solution supplied by one of the Chinese contestants after the
competition. It requires only one, although tricky, use of the AM-GM inequality.
n
X
i=1
xi
4
=
n
X
x2i
i=1
+2
X
xi xj
1≤i<j≤n
2
n
X
X
2
≥4
xk 2
xi xj =
k=1
1≤i<j≤n
n
X
X
X
2
8
xk ≥ 8
xi xj (x2i + x2j ).
xi xj
1≤i<j≤n
1≤i<j≤n
k=1
The second inequality is an equality if and only if n − 2 of the xi ’s are zeros. Let us therefore
then assume that x3 = x4 = · · · = xn = 0. Then, for the first inequality to become an equality
one requires that (x21 + x22 + 2x1 x2 )2 = 8(x21 + x22 )x1 x2 , which reduces to (x1 − x2 )4 = 0. Hence
1
x1 = x2 . The answer is C = with equality if and only if two of the xi ’s are equal and the rest
8
are zeros.
2. Chebyshev’s inequality
Let a1 , a2 , ..., an and b1 , b2 , ..., bn be two sequences of real numbers, at least one of which consists
entirely of positive numbers. Assume that a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn . Then
1
1
1
(a1 + a2 + · · · + an ) · (b1 + b2 + · · · + bn ) ≤ (a1 b1 + a2 b2 + · · · + an bn ).
n
n
n
If we instead assume that a1 ≤ a2 ≤ · · · ≤ an and b1 ≥ b2 ≥ · · · ≥ bn , then the inequality is
reversed. Equality occurs if and only if at least one of the sequences is constant.
3
Example 5. Let a1 , a2 , ..., an be a sequence of positive real numbers with the arithmetical mean
n
1 2 1 X
1 2
ak +
A. Prove that
≥ A+
.
n k=1
ak
A
n
n
1 2
1 X 2 1 X 1 2
Solution. By squaring we get the equivalent inequality
≥A +
a +
.
n k=1 k n k=1 a2k
A
We will show it in two steps.
We can, without loss of generality, assume that the sequence a1 , a2 , ..., an is increasing. Then
the sequence 1/a1 , 1/a2 , ..., 1/an is decreasing and, by using Chebyshev’s inequality twice, we get
n
1 X 1 2
A =
n k=1 a2k
n
n
n
n
n
n
1 X 1 1 X 1 X 1 X 1 1 X 1 X
ak
ak ≥
ak ≥
1 = 1.
n k=1 a2k n k=1
n k=1
n k=1 ak n k=1
n k=1
n
n
n
1 X 2 1 X 1 X From Chebyshev’s inequality we get also
a ≥
ak
ak
n k=1 k
n k=1
n k=1
= A2 .
These two results together give the inequality in question.
Example 6. Let b1 , b2 , ..., bn be positive real numbers. Show that
n
n
X
1 2 X 2
bi ≥ n 3 .
b
i=1
i=1 i
n
n
n
h X
1 i h X 1 i h X 2 i
Solution. We may write the desired inequality as
/n ·
/n ·
bi /n ≥
b
b
i
i
i=1
i=1
i=1
1. Because of the symmetry we may assume that the sequence b1 , b2 , ..., bn is increasing and
1 1
1
then we notice that the sequence , , ...,
is decreasing. Using Chebyshev’s inequality twice,
b1 b2
bn
n
n
n
h X
1 i h X 1 i h X 2 i
first with the second and the third factor, we get
/n ·
/n ·
bi /n ≥
b
b
i=1 i
i=1 i
i=1
n
n
n
h X
X
1 i h X i
/n ·
bi /n ≥
1 /n = 1.
b
i=1 i
i=1
i=1
Example 7. (India, 1995) Let n be an integer greater than 1 and let a1 , a2 , ..., an be positive
real numbers with the sum 1. Show that
r
a2
an
n
a1
√
+√
+ ··· + √
≥
.
n−1
1 − a1
1 − a2
1 − an
Solution. We may assume that the sequence a1 , a2 , ..., an is increasing and then the sequence
a1
is increasing as well. Chebyshev’s inequality now implies that √
+
1 − a1
1
1
√ 1
, √1−a
, ..., √1−a
1−a1
n
2
4
a2
an
1
1
1
1
+ ··· + √
≥ (a1 + a2 + ... + an ) √
+√
+ ··· + √
=
n
1 − a2
1 − an
1 − a1
1 − a2
1 − an
1
1
1
1
√
+√
+ ··· + √
.
n 1 − a1
1 − a2
1 − an
1
, we get
Now we can use the result from the previous example. Letting bi = √
1 − ai
n
n
n
−1
X
−1
1 X
2
X
n
2
=n
(1 − ai )
=
bi ≥ n
bi
.
n i=1
n−1
i=1
i=1
r
1
1
1
1
n
√
Hence
+√
+ ··· + √
≥
.
n 1 − a1
n−1
1 − a2
1 − an
1
The equality holds if and only if a1 = a2 = · · · = an = .
n
√
3. Rearrangement inequality
Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be real numbers. For any permutation
0
0
0
(a1 , a2 , ..., an ) of (a1 , a2 , ..., an ), we have
0
0
0
a1 b1 + a2 b2 + · · · + an bn ≥ a1 b1 + a2 b2 + · · · + an bn ≥ an b1 + an−1 b2 + · · · + aa bn ,
0
0
0
with equality if and only if the sequence (a1 , a2 , ..., an ) is equal to (a1 , a2 , ..., an ) or (an , an−1 , ..., a1 )
respectively.
a1 a2
With the notation
b1 b2
above may be written as
0
a1 a2 ... an
a1
≥
b1 b2 ... bn
b1
... an
... bn
0
= a1 b1 + a2 b2 + · · · an bn the statement of the inequality
0
a2 ... an
b2 ... bn
≥
an an−1 ... a1
b1
b2 ... bn
This innocent-looking and easy to prove inequality is in fact a very powerful instrument and
several other inequalities here can easily be derived from it. The Rearrangemant inequality is one
of my two favourites. The second one is the Muirhead’s inequality.
Example 8. (IMO 1975) Let xi , yi , 1 ≤ i ≤ n, be real numbers such that x1 ≥ x2 ≥ · · · ≥
xn and y1 ≥ y2 ≥ · · · ≥ yn . Let z1 , z2 , ..., zn be any permutation of y1 , y2 , ..., yn . Show that
n
n
X
X
2
(xi − yi ) ≤
(xi − zi )2 .
i=1
i=1
Solution. After squaring and cancelling equal terms, the inequality becomes
n
X
i=1
which basically is the Rearrangement inequality.
5
xi yi ≥
n
X
xi zi
i=1
Example 9. For all positive numbers a, b and c, prove the inequality
a3 b a3 c b 3 a b 3 c c 3 a c 3 b
+
+
+
+
+
≥ 6abc
c
b
c
a
b
a
Solution. Because of the symmetry we may assume that a ≥ b ≥ c. Then
2 2 2 2 2 2 a3 b b 3 c c 3 a
ab bc ca
≥
the Rearrangement inequality implies that
+
+
=
a
b
c
c
a
b
bc
ca
ab
2 2 2 2 2 2 2 2 2 2 2 2 a3 c b 3 a c 3 b
ab bc ca
ab bc ca
= 3abc and, similarly,
≥
+
+
=
c
a
b
c
a
b
b
c
a
ab
bc
ca
ca
ab
bc
2 2 2 2 2 2 ab bc ca
= 3abc. Adding these two inequalities together yields the desired result. c
a
b
ab
bc
ca
Example 10. For all positive numbers a, b and c prove the inequality
a2
b2
c2
a+b+c
+
+
≥
.
b+c c+a a+b
2
Solution. Because of the symmetry we may assume a ≥ b ≥ c. Then
2
2
2
a2 b 2 c 2
a
b2 c2
a
b2 c 2
a
≥
and
≥
1
1
1
1
1
1
1
1
1
1
b+c
c+a
a+b
c+a
a+b
b+c
b+c
c+a
a+b
a+b
b2
c2
1
b+c
1
c+a
After adding those two inequalities and using the easy to prove inequality
for positive x, y, the result follows.
.
x+y
x2 + y 2
≥
,
x+y
2
Example 11. (IMO, 1983) Let a, b, c be the sides of a triangle. Prove that
a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0.
Solution. We may assume a ≥ max{b, c}. If a ≥ b ≥ c, we first prove that a(b + c − a) ≤
b(c + a − b) ≤ c(a + b − c). Note that b(c + a − b) − a(b + c − a) = (a − b)(a + b − c) ≥ 0. The
second inequality reduces to (b − c)(b + c − a) ≥ 0 and is obvious by the triangle inequality.
1
1
1
Dividing the desired inequality by abc we have to show that a(a−b)+ b(b−c)+ c(c−a) ≥
c
a
b
1
1
1
0, or, after subtracting a+b+c, that a(−c+a−b)+ b(−a+b−c)+ c(−b+c−a) ≥ −(a+b+c).
c
a
b
1
1
1
The desired inequality can be written as a(c − a + b) + b(a − b + c) + c(b − c + a) ≤ a + b + c.
c
a
b
1
1
1
By the Rearrangement inequality a(c − a + b) + b(a − b + c) + c(b − c + a) =
c
a
b
a(c − a + b) b(a − b + c) c(b − c + a)
≤
1
1
1
c
a
b
6
a(c − a + b) b(a − b + c) c(b − c + a)
1
a
1
b
1
c
= a + b + c.
If a ≥ c ≥ b the proof is similar.
4. Cauchy-Schwarz inequality
For any real numbers a1 , a2 , ..., an and b1 , b2 , ..., bn ,
(a1 b1 + a2 b2 + · · · + an bn )2 ≤ (a21 + a22 + · · · + a2n ) · (b21 + b22 + · · · + b2n ),
with equality if and only if there are two real numbers α and β, not both equal to 0, such that
αak = βbk for all k = 1, 2, ..., n.
1 1 1
+ + = 2. Prove that
x y z
p
√
√
√
x + y + z ≥ x − 1 + y − 1 + z − 1.
Example 12. (Iran, 1998) Let x, y, z > 1 and
Solution. Observe that by hypothesis
x−1 y−1 z−1
+
+
= 1. Then, by using the Cauchyx
y
z
Schwarz inequality, we get
r
r
r
2
2
2
p
√
√ 2 √ 2 √ 2
x−1
y−1
z−1 x+y+z =
x + y + z
+
+
x − 1+ y − 1+
≥
x
y
z
√
2
z − 1 , which gives the desired inequality.
Example 13. (Romania, 1999) Let n ≥ 2 be a positive integer and let x1 , x2 , ..., xn , y1 , y2 , ..., yn
be positive real numbers such that x1 + x2 + · · · + xn ≥ x1 y1 + x2 y2 + · · · xn yn . Prove that
x1 x2
xn
x1 + x2 + · · · + xn ≤
+
+ ··· .
y1
y2
yn
Solution. Applying the Cauchy-Schwarz inequality and then the given inequality, we have
n
n
n
n
n
n
X
2 X
X
X
X
X
xi
xi
xi ≤
xi yi ·
≤
xi ·
. Dividing both sides by
xi yields the desired
y
y
i
i
i=1
i=1
i=1
i=1
i=1
i=1
inequality.
Example 14. (USSR, 1986) Let a1 , a2 , ..., an be positive real numbers. Prove that
2
3
n
1
1
1
1
+
+
+ ··· +
<2
+
+ ··· +
.
a1 a1 + a2 a1 + a2 + a3
a1 + a2 + · · · + an
a1 a2
an
Solution. Let Sk =
k
X
i=1
ai and Ak =
k
X
i2
, for k = 1, 2, ..., n. Using the Cauchy-Schwarz
a
i
i=1
7
k
k
k
X
i √ 2 X i2 X
√ · ai ≤
inequality we now get
·
=
a i = Ak S k .
2
a
a
i
i
i=1
i=1
i=1
k
4kAk
(4k + 2)Ak 2
2
Hence,
≤ 2
Ak . By adding the left-hand sides
< 2
=
−
Sk
k (k + 1)2
k (k + 1)2
k 2 (k + 1)2
for all k, we receive
n
n
n
n+1
n
X
X
X
X
X
k
2
2
2
2
2
1
≤
+
−
Ak −
Ak−1 =
(Ak − Ak−1 ) −
An = 2
2
2
2
2
Sk
k
k
a1 k=2 k
(n + 1)
ak
k=1
k=1
k=2
k=1
n
X
2
1
.
An < 2
2
(n + 1)
ak
k=1
k(k + 1) 2
5. Hölder’s inequality
Let p and q be two positive real numbers whose sum is 1 and let a1 , a2 , ..., an and b1 , b2 , ..., bn be
two sequences of positive real numbers. Then
n
X
ak b k ≤
n
X
k=1
1/p
ak
n
p X
q
1/q
·
bk
,
k=1
k=1
with the equality if and only if there are two real numbers α and β, not both equal to 0, such that
1/p
1/q
αak = βbk for all k = 1, 2, ..., n.
This inequality can easily be generalized to more than two sequences:
Suppose {ai1 , ai2 , ..., ain }, i = 1, 2, ..., k, are k sequences of positive real numbers and
p1 , p2 , ..., pk are positive real numbers whose sum is 1. Then,
n
X
j=1
a1j a2j ...akj ≤
n
X
1/p1
a1j
n
n
p1 X
p2
X
pk
1/p
1/p
·
a2j 2
· ... ·
akj k .
j=1
j=1
j=1
Example 15. (Belarus, 2000) For all positive real numbers a, b, c, x, y, z prove that
(a + b + c)3
a3 b 3 c 3
+ +
≥
.
x
y
z
3(x + y + z)
Solution. We use the Hölder inequality generalized to three sequences: p1 p2 p3 + q1 q2 q3 +
3
Y
r 1 r2 r3 ≤
(p3i + qi3 + ri3 )1/3 , for all positive real numbers pi , qi , ri .
i=1
a3
b3 c3 1/3
Hence,
+ +
(1 + 1 + 1)1/3 (x + y + z)1/3 ≥ a + b + c. Cubing both sides and
x
y
z
then dividing by 3(x + y + z) give the desired result.
8
Example 16. Let a, b, c, d be positive real numbers. Prove that
a6 b3 + b6 c3 + c6 d3 + d6 a3 ≥ a2 b5 c2 + b2 c5 d2 + c2 d5 a2 + d2 a5 b2 .
Solution. Let x = (a2 b)3 , y = (b2 c)3 , z = (c2 d)3 and t = (d2 a)3 . Then a6 b3 + b6 c3 + c6 d3 +
3
d6 a3 = (x + y + z + t)1/3 = (x + y + z + t)1/3 (y + z + t + x)1/3 (y + z + t + x)1/3 ≥
1/3
1/3
1/3
1/3
= a2 b5 c2 + b2 c5 d2 + c2 d5 a2 + d2 a5 b2 .
+ tx2
+ zt2
+ yz 2
xy 2
6. Minkowski’s inequality
Given a real number r ≥ 1 and two sequences of positive real numbers, a1 , a2 , ..., an and b1 , b2 , ..., bn ,
we have
n
n
n
X
1/r X
1/r X
1/r
r
r
(ai + bi )
≤
ai
+
bri
,
i=1
i=1
i=1
with equality if and only if there are two real numbers α and β, not both equal 0, such that αak =
βbk for all k = 1, 2, ..., n.
For 0 < r < 1 the inequality is reversed.
It is again obvious that this inequality may be generalized to more than two sequences of
positive real numbers.
Example 17. For all nonnegative real numbers x, y, z prove that
√ √
√
√
√
√
√
x + y + y + z + z + x ≥ 2( x + y + z).
√
√
√
Solution. By taking r = 2 and letting a1 = x, a2 = y, a3 = z we may
use Minkowski’s
2
1/2
1/2
1/2
inequality in an obvious way: a21 + a22
+ a22 + a23
+ a23 + a21
≥ a1 + a2 + a3 +
2 1/2 √
a2 + a3 + a1
= 2(a1 + a2 + a3 ).
Example 18. Let a1 , a2 , ..., an be a sequence of positive real numbers with the product a1 a2 · · · an =
1. Suppose that (b1 , b2 , ..., bn ), (c1 , c2 , ..., cn ) and (d1 , d2 , ..., dn ) are three permutations of the sequence (a1 , a2 , ..., an ). Prove that
n p
X
ak + bk + ck + dk ≥ 2n.
k=1
Solution. By Minkowski’s inequality
n p
X
ak + bk + ck + dk
k=1
2
≥
n
n
X
√ 2 X p 2
ak +
bk +
k=1
9
k=1
n
n
n
X
X
√ 2 X p 2
√ 2
ck +
dk = 4
ak .
k=1
k=1
By the AM-GM inequality
k=1
n
X
√
√ √
√
a1 a2 · · · an =
q
ak ≥ n n
k=1
n
√
2n
a1 a2 · · · an = n. Those two inequalities together imply
n p
X
ak + bk + ck + dk ≥ 2n.
k=1
7. Jensen’s inequality
Let f (x) be a strictly convex function on an interval I and let α1 , α2 , ..., αn be nonnegative numbers
n
X
such that
αk = 1. Then, for all x1 , x2 , ..., xn ∈ I,
k=1
f (α1 x1 + α2 x2 + · · · + αn xn ) ≤ α1 f (x1 ) + α2 f (x2 ) + · · · + αn f (xn ),
with equality if and only if x1 = x2 = · · · = xn .
In the case when f (x) is strictly concave on I, the inequality is reversed.
The convexity/concavity of a function f (x) may be checked using derivative tests:
(1) Let f (x) be a differentiable function on an interval I. Then f (x) is strictly convex
(concave) on I, if and only if f 0 (x) is strictly increasing (decreasing) on I.
(2) Let f (x) be a twice differentiable function on an interval I. Then f (x) is strictly
convex (concave) on I, if and only if f 00 (x) > 0 (f 00 (x) < 0) for all x in the interior of I.
Example 19. (same as Example 7 above) Let n be an integer greater than 1 and let a1 , a2 , ..., an
a2
an
a1
+√
+ ··· + √
≥
be positive real numbers with the sum 1. Show that √
1
−
a
1
−
a
1
−
a
1
2
n
r
n
.
n−1
Solution. Since all ai in question are from the interval I = (0, 1) we may consider the function
x
4−x
f (x) = √
which is differentiable on I. Calculation gives f 00 (x) =
. It is then
4(1 − x)5/2
1−x
obvious that f 00 (x) is positive on I, and therefore strictly convex there.
a1
a2
an
Hence we may use Jensen’s inequality: √
+√
+ ··· + √
1 − a1
1 − a2
1 − ar
n
f (a1 ) + f (a2 ) + · · · + f (an )
a1 + a2 + · · · + an 1
n
=n·
≥n·f
=n·f
=
.
n
n
n
n−1
Example 20. (Korea, 1998) For positive real numbers a, b, c with
10
a + b + c = abc, show that
√
1
1
1
3
+√
+√
≤
2
2
2
2
1+a
1+b
1+c
and determine when equality occurs.
Solution. The formulas on the left-hand side suggest the substitution α = arctan a, β =
arctan b and γ = arctan c. This substitution together with the condition tan α + tan β + tan γ =
tan α tan β tan γ implie that 0 < α, β, γ < π2 and α + β + γ = π. We want to show that
3
cos α + cos β + cos γ ≤ .
2
Taking f (x) = cos x we notice that f 00 (x) = − cos x < 0 for 0 < x < π, hence f (x) is strictly
concave inside the interval (0, π2 ).
1
1
Now we can use Jensen’s inequality: (cos α + cos β + cos γ) = f (α) + f (β) + f (γ) ≤
3
3
α + β + γ
π
1
f
= cos = .
3
3
2
Example 21. (USA, 1974) For positive real numbers a, b, c prove that
aa bb cc ≥ (abc)(a+b+c)/3 .
Solution. Consider the function f (x) = ln xx = x ln x. We have f 00 (x) =
for x > 0. Hence f (x) is strictly convex for x > 0.
Now, applying Jensen’s inequality we get
1
which is positive
x
1
ln aa + ln bb + ln cc
ln aa + ln bb + ln cc
f (a) + f (b) + f (c)
ln(aa bb cc ) =
=
=
3
3
3
3
(a+b+c)/3
a+b+c
a+b+c
= ln
.
≥f
3
3
1/3
a+b+c
From the AM-GM inequality we have
≥ abc
. Hence,
3
a + b + c (a+b+c)/3
(a+b+c)/9
≥ abc
and, since logarithm is an increasing function, then we get
3
a +
(a+b+c)/3
(a+b+c)/9 1
(a+b+c)/3
b+c
ln
≥ ln abc
= ln abc
.
3
3
From these two results the desired inequality follows.
8. Power Mean inequality
Let a1 , a2 , ..., an be nonnegative real numbers, k and m positive real numbers with k ≤ m. Then
1/k 1
1/m
1
m
m
≤
,
(ak1 + ak2 + · · · akn )
(am
+
a
+
·
·
·
a
)
2
n
n
n 1
11
with equality if and only if a1 = a2 = · · · = an .
Example 22. (North African MO, 1986) Let a1 , a2 , a3 be positive real numbers. Prove that
1
1 1
1
1 2
1
+
+
≥4
+
+
.
3
a1 a2 a2 a3 a3 a1
a1 + a2 a2 + a3 a3 + a1
√
a1 + a2
4
1
≥ a1 a2 implies
. Simi≤
2
2
(a1 + a2 )
a1 a2
1
1
1 lar inequalities for the other two terms of the left-hand-side give 3
+
+
≥
a1 a2
a2 a3
a3 a1
1
1
1
.
12
+
+
(a1 + a2 )2 (a2 + a3 )2 (a3 + a1 )2
1
The rest follows from the Power Mean inequality (n = 3, m = 2 and k = 1): 12
+
(a1 + a2 )2
1
1
1
1
1
1
=
36
/3
≥
36
(
+
+
+
+
2
2
2
2
2
(a2 + a3 )
(a3 + a1 )
(a1 + a2 )
(a2 + a3 )
(a3 + a1 )
a1 + a2
1
1 2
1
1 2
1
+
/3 = 4
+
+
.
a2 + a3 a3 + a1
a1 + a2 a2 + a3 a3 + a1
Solution. The AM-GM inequality
Example 23. (Czech Republic and Slovakia, 2000) Show that
r
r
r
1 1
a
b
3
3
3
+
≤ 2(a + b) +
b
a
a b
for all positive real numbers a and b, and determine when equality occurs.
b 1/6
a 1/6
and a2 =
), we have
Solution. By the Power Mean inequality (with a1 =
b
a
q
q
p
p
3 ab + 3 ab 3 ab + ab 2
≤
, with equality if and only if a/b = b/a, or equivalently a = b.
2
2
pa q b
b + a 2 a + b 1 1 The desired result follows from the identity
=
+ .
2
4 a b
Example 24. Let x, y, z be positive real numbers. Show that
x6
y6
z6
x5 + y 5 + z 5 ≤ √ + √ + √ .
yz
xy
xz
√
√
√
Solution. Letting a = x, b = y, c = z and clearing the denominators we get the equivalent inequality (a10 + b10 + c10 )abc ≤ a13 + b13 + c13 . Now,
r a13 + b13 + c13 10 r a13 + b13 + c13 3
r a13 + b13 + c13 13
13
13
13
a13 +b13 +c13 = 3
=3
≥
3
3
3
√ 3
r a10 + b10 + c10 10 a + b + c 3
10
3
3
≥ (a10 + b10 + c10 ) abc = (a10 + b10 + c10 )abc. 3
3
12
9. Schur’s inequality
Let x, y and z be nonnegative real numbers. Then for any r > 0,
xr (x − y)(x − z) + y r (y − z)(y − x) + z r (z − x)(z − y) ≥ 0,
with equality if and only if x = y = z or if two of x, y, z are equal and the third is 0.
In the case of the r = 1 this inequality is often written equivalently as x3 + y 3 + z 3 + 3xyz ≥
x2 y + xy 2 + y 2 z + yz 2 + z 2 x + zx2 .
In the next example we use another trick, homogenization, which is very useful when dealing
with polynomial inequalities, provided there is an additional constraint, like x + y + z = 1 or
xyz = 1. One may then multiply some terms of the inequality with the constraint in order to get
all terms of the same degree.
For example, the inequality x2 y + xz ≤ 2z + 7 with xyz = 1 (for positive real numbers x, y, z),
is equivalent to the homogeneous inequality x2 y + xz(xyz)1/3 ≤ 2z(xyz)2/3 + 7(xyz), where each
term now has degree 3.
Now, applying the substitution x = u3 , y = v 3 and z = w3 and dividing by common factors
we end up with the equivalent inequality u4 v 2 + u2 w4 ≤ 2vw5 + 7uv 2 w3 which may be easier to
deal with.
Example 25. (IMO, 1984) Let x, y, z be non negative real numbers such that x + y + z = 1.
Prove that
7
0 ≤ xy + yz + zx − 2xyz ≤ .
27
Solution. Using the condition x + y + z = 1, we can reduce the given inequality to a homoge7
neous one, 0 ≤ (xy + yz + zx)(x + y + z) − 2xyz ≤ (x + y + z)3 . The first inequality reduces
27
to 0 ≤ xyz + x2 y + xy 2 + y 2 z + yz 2 + z 2 x + zx2 , which is obviously true.
The second inequality simplifies to 7(x3 + y 3 + z 3 ) + 15xyz ≥ 6(x2 y + xy 2 + y 2 z + yz 2 +
z x + zx2 ). From the AM-GM inequality we can deduce that x3 + y 3 + z 3 ≥ 3xyz. Hence,
using Schur’s inequality for r = 1, 7(x3 + y 3 + z 3 ) + 15xyz ≥ 6(x3 + y 3 + z 3 ) + 18xyz ≥
6(x2 y + xy 2 + y 2 z + yz 2 + z 2 x + zx2 ).
2
Example 26. (IMO, 2000) Let a, b, c be positive real numbers such that abc = 1. Prove that
1 1 1
a−1+
b−1+
c−1+
≤ 1.
b
c
a
Solution. The expression is equivalent to the following homogeneous inequality: a−(abc)1/3 +
(abc)2/3 (abc)2/3 (abc)2/3 b − (abc)1/3 +
c − (abc)1/3 +
≤ abc.
b
c
a
After the substitution a = x3 , b = y 3 , c = z 3 , (x, y, z > 0), the inequality becomes (x2 y −
y 2 z + z 2 x)(y 2 z − z 2 x + x2 y)(z 2 x − x2 y + y 2 z) ≤ x3 y 3 z 3 .
13
Again, using the substitution x2 y = u, y 2 z = v, z 2 x = w, the inequality can be written as
3uvw + (u3 + v 3 + w3 ) ≥ u2 v + v 2 u + v 2 w + w2 v + w2 u + u2 w. And this exactly is what Schur’s
inequality for r = 1 says.
10. Maclaurin’s inequality
Let a1 , a2 , ..., an be positive real numbers. Then,
h 1 X i1 h 1 X
i1/2 h 1 X
i1/3
a
a
a
a
a
a
≥
≥
≥ ··· ≥
i
i j
i j k
n
n
n
1
i
2
3
i<j
i<j<k
i1/n
h 1
,
n a1 a2 · · · an
n
with equality if and only if a1 = a2 = · · · = an .
This is a very elegant generalization of the AM-GM inequality.
With the notation Sk = the sum
of all products of k-subsets of {a1 , a2 , ..., an } divided by nk (for k = 1, 2, ..., n) we can write the
Maclaurin’s inequality as
1/2
AM = S1 ≥ S2
1/3
≥ S3
≥ · · · ≥ Sn1/n = GM.
Example 27. (Poland, 1989) Suppose a, b, c, d are positive real numbers. Show that
r
r
ab + ac + ad + bc + bd + cd
3 abc + abd + acd + bcd
≥
.
6
4
Solution. This is only a special case of Maclaurin’s inequality: the second and the third expression for n = 4.
Example 28. Let a, b, c be positive real numbers. Show that
1 1 1
a8 + b 8 + c 8
≥ + + .
3
3
3
abc
a b c
Solution. Although it is a rather easy exercise when using the AM-GM-HM inequality we will
a + b + c 8
a8 + b 8 + c 8
thy it bu more sofisticated method. By the Power Mean inequality
≥
.
3
a + b + c 8 a + b + c 6 a + b3+ c 2
ab + bc + ca
Now, using Maclaurin’s inequality,
=
≥ (abc)2 ·
.
3
3
3
3
a8 + b 8 + c 8
3
a + b + c 8 ab + bc + ac
1 1 1
Hence,
≥
≥
= + + .
3
3
3
3
abc
(abc)
3
abc
a b c
14
11. Muirhead’s inequality
In order to simplify the notation, let us introduce the symmetric summation symbol
X
. Let
sym
P (x, y, z) be a function of three variables, x, y and z. Let us define
X
P (x, y, z) = P (x, y, z) +
sym
P (x, z, y) + P (y, x, z) + P (y, z, x) + P (z, x, y) + P (z, y, x).
For example, for P1 (x, y, z) = x3 , P2 (x, y, z) = x2 y 2 z 2 , P3 (x, y, z) = x3 y 2 , we have
X
X
X
P1 (x, y, z) = 2x3 + 2y 3 + 2z 3 ,
P2 (x, y, z) = 6x2 y 2 z 2 and
P3 (x, y, z) = x3 y 2 +
sym
2 3
sym
3 2
2 3
3 2
sym
2 3
x y +x z +x z +y z +y z .
This notation may of course be generalized to any number n ≥ 1 variables, but for our purposes
n = 3 will be sufficient.
Muirhead’s inequality for three variables may be now stated as follows:
Given six non negative real numbers a1 , a2 , a3 , b1 , b2 , b3 such that
a1 ≥ a2 ≥ a3 , b 1 ≥ b 2 ≥ b 3 ,
a1 ≥ b1 , a1 + a2 ≥ b1 + b2 and a1 + a2 + a3 = b1 + b2 + b3
then, for all non negative real numbers x, y, z
X
X
x a1 y a2 z a3 ≥
xb1 y b2 z b3 .
sym
sym
This particular inequality may turn out to be very useful when many other solving methods fail.
Using it requires however that we deal with a homogeneous inequality.
Example 29. (USA, 1997) Prove that, for all positive real numbers a, b, c,
a3
1
1
1
1
+ 3
+ 3
≤
.
3
3
3
+ b + abc b + c + abc c + a + abc
abc
Solution. Clearing the denominators gives a rather uninviting expression (a3 + b3 + abc)(b3 +
c3 + abc) + (a3 + b3 + abc)(c3 + a3 + abc) + (b3 + c3 + abc)(c3 + a3 + abc) abc ≤ (a3 + b3 +
abc)(b3 + c3 + abc)(c3 + a3 + abc). The trick is however to multiply both sides with 2, because we
may then see the left-hand side as a symmetric sum and write the inequality as
X
(a3 + b3 + abc)(a3 + c3 + abc)abc ≤ 2(a3 + b3 + abc)(b3 + c3 + abc)(c3 + a3 + abc).
sym
This may be simplified to
X
X
(a7 bc + 3a4 b4 c + 4a5 b2 c2 + a3 b3 c3 ) ≤
(a7 bc + 2a6 b3 + 2a5 b2 c2 + 3a4 b4 c + a3 b3 c3 ).
sym
sym
15
The last inequality reduces to
X
a6 b3 − a5 b2 c2 ≥ 0, which is clearly true by Muirhead’s inequa-
sym
lity (with a1 = 6, a2 = 3, a3 = 0, b1 = 5, b2 = 2, b3 = 2).
Example 30. (IMO, 1995) Prove the following inequality for positive real numbers x, y, z, such
that xyz = 1:
1
1
3
1
+ 3
+ 3
≥ .
3
x (y + z) y (z + x) z (x + y)
2
Solution. Suppose that we find no other ”clever” way of solving the problem and the only
methods we can apply is by direct counting.
We start with making the inequality homogenous by multiplying the denominator of the righthand side by (xyz)4/3 . Then, in order to make our calculations easier we make the substitution
x = a3 , y = b3 and z = c3 , with a, b, c > 0. The inequality is then equivalent to
1
1
1
3
+ 9 3
+ 9 3
≥ 4 4 4.
3
3
3
+ c ) b (c + a ) c (a + b )
2a b c
a9 (b3
Clearing the denominators leads to the following inequality (it should be mentioned that you don’t
have to first multiply all terms in order to find the appropriate symmetric sum. After practising
with some exercises you will be able to quickly discover the pattern and find those sums.)
X
X
(3a11 b8 c5 + a8 b8 c8 ).
(a12 b12 + 2a12 b9 c3 + a9 b9 c6 ) ≥
sym
sym
This may in turn be reduced to
X
X
X 9 9 6
a12 b12 − a11 b8 c5 + 2
a12 b9 c3 − a11 b8 c5 +
a b c − a8 b8 c8 ≥ 0.
sym
sym
sym
The Muirhead’s inequality implies now that every term on the left-hand side is ≥ 0.
There are of course many other ways to solve this problem. One is to use the Cauchy-Schwarz
x2
+
inequality. We may substitute x = a1 , y = 1b and z = 1c , and then the inequality becomes
y+z
y2
z2
3
+
≥ .
x+z x+y
2
√
√
√
x
y
z
Now, since x + y + z = √
· y+z+ √
· x+z+ √
· x + y, then, by the
y+z
x+y
x+z
2
2
2 x
y
z
Cauchy-Schwarz inequality (x + y + z)2 ≤
+
+
(y + z) + (x + z) + (x + y) .
y+z x+z x+y
x2
y2
z2
x+y+z
3√
3
Hence,
+
+
≥
≥ 3 xyz = . The last inequality follows from
y+z
x+z
x+y
2
2
2
the AM-GM inequality.
16
Yet another way to approach the problem is by using the Rearrangement inequality. Because the inequality we want to show is symmetric, and using the same substitution as in the previous solution,
1
1
1
we may assume that x ≥ y ≥ z. Then x2 ≥ y 2 ≥ z 2 and y+z
≥ x+z
≥ x+y
. Now, since
2
2
2
2
x
y2 z2
x
y2 z2
x
y2 z2
x
y2 z2
≥
and
≥
1
1
1
1
1
1
1
1
1
1
1
1
y+z
x+z
x+y
x+y
y+z
x+z
y+z
x+z
x+y
x+z
x+y
y+z
then, after adding those two inequalities and dividing by 2, we get
y2
z2
1 x2 + y 2 y 2 + z 2 x2 + z 2 x2
+
+
≥
+
+
.
y+z x+z x+y
2 x+y
y+z
x+z
1 x + y y + z x + z
Since s2 + t2 ≥ (s + t)2 /2 then the right hand side is ≥
+
+
=
2
2
2
2
x+y+z
3
x+y+z
3√
. Finally, the AM-GM inequality implies that
≥ 3 xyz = .
2
2
2
2
12. Substitutions
The process of treating inequalities may sometimes be simplified by suitable substitutions, as we
already have seen before in Example 20.
Example 31. (Russia, 2000) For real numbers x, y such that 0 ≤ x, y ≤ 1, prove that
√
2
1
1
≤√
+p
.
1 + xy
1 + x2
1 + y2
1
≤ 2, which is obviously
1 + y2
true. So, suppose 0 < x, y ≤ 1. The inequality looks almost like a candidate for Jensen’s inequality
1
with the concave function f (x) = √
. The only obstacle is the right-hand product xy instead
1 + x2
of the sum x + y.
1
.
Because of that it seems to be appropriate to instead study the function g(s) = √
1 + e−2s
To that end we make the substitutions x = e−u and y = e−v for some nonnegative real num1
1
bers u and v. The problem then reduces to the equivalent inequality √
+√
≤
−2u
1+e
1 + e−2v
2
√
.
1 + e−(u+v)
To finish the solution, all we need is to show that the function g(s) above is concave for s ≥ 0.
1 − 2e2s
Easy calculations give g 00 (s) =
. The denominator is certainly positive, while the
(1 + e−2s )5/2 e4s
numerator is negative, because e2s ≥ 1 for s ≥ 0. Thus, g(s) is strictly concave for s ≥ 0 and the
solution is complete.
Solution. If x = 0, the problem reduces to the inequality 1 + p
17
Example 32. Prove that, for all positive real numbers a, b, c, d,
p
√
√
ab + cd ≤ (a + d)(b + c).
Solution. Dividing
the inequality
r with the expression on the right-hand side we get the equivar
a
b
c
d
·
+
·
≤ 1, or
lent inequality
a+d b+c
b+c a+d
r
r
a
b b
a 1−
·
+
· 1−
≤1
a+d b+c
b+c
a+d
b
a
= sin2 α and
= cos2 β for some α, β, 0 < α, β < π2 , the left-hand side
Letting
a+d
b+c
can be written as sin α sin β + cos α cos β = cos(α − β), which obviously is ≤ 1.
PROBLEMS 1
With each problem comes my suggestion for a strategy of solving it. Such a strategy is never unique
and these problems may certainly be solved by alternative methods as well.
1. (St. Petersburg, 1997) Prove that for x, y, z ≥ 2, (x + y 3 )(y + z 3 )(z + x3 ) ≥ 125xyz.
(Hint: x + y 3 ≥ x + 4y. Now, apply the AM-GM inequality.)
x
y
1
+ 2
≤
.
2
4
+y
x +y
xy
(Hint: Apply the AM-GM inequality to denominators..)
2. (Russia, 1995) Prove that for x, y > 0,
x4
3. Let a, b, c, d be positive real numbers. Prove that
(Hint: the AM-GM inequality.)
1 4 4 16
64
+ + +
≥
.
a b c
d
a+b+c+d
a 4. (Asian-Pacific MO, 1998) Let a, b, c be positive real numbers. Prove that 1 +
1+
b
b c
a + b + c
1+
≥2 1+ √
.
3
c
a
abc
(Hint: the AM-GM inequality twice.)
5. (Poland, 1990) Let√x, y, z√be positive
real numbers such that xyz = 2. Show that x2 + y 2 +
√
z + xy + yz + xz ≥ 2( x + y + z).
(Hint: You may start by showing that x2 + y 2 + z 2 ≥ xy + yz + zx and then use the AM-GM
inequality.)
2
18
6. (Vietnam, 1998) Let n be an integer, n ≥ 2, and let x1 , x2 , ..., xn be
positive numbers satis√
n
x1 x2 · · · xn
1
1
1
1
+
+ ··· +
=
. Prove that
≥ 1998.
fying
x1 + 1998 x2 + 1998
xn + 1998
1998
n−1
X
1998
observe that 1 − yi =
(Hint: For yi =
yk . Now use the AM-GM inequality.)
xi + 1998
k6=i
7. (Belarus, 1999) Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that
1
1
1
3
+
+
≥ .
1 + ab 1 + bc 1 + ca
2
(Hint: use the AM-HM inequality and the inequality a2 + b2 + c2 ≥ ab + bc + ca.)
8. (St. Petersburg, 1999) Let x0 > x1 > · · · > xn be real numbers. Prove that x0 +
1
1
+ ··· +
≥ xn + 2n.
x1 − x2
xn−1 − xn
(Hint: the AM-GM inequality for t +
1
+
x0 − x1
1
≥ 2.)
t
64x2 y 2
= (x + 1)(y + 2)(2x + y).
4x2 + y 2
(Hint: After clearing the denominator use the AM-GM inequality on the right-hand side expression.)
9. Find all pairs of positive real numbers x, y such that
10. (Poland, 1990) Let a, b be two positive real numbers. Find all pairs of positive real numbers
27xy
1
x y
x, y such that
=
+ + .
(1 + 2ax)(1 + 2by)
ab a b
(Hint: One may use the AM-GM inequality three times: for 1 + 2ax, for 1 + 2by and for the
right hand side of the equality.)
11. Let n be a positive integer and let a, b, c be positive real numbers. Show the inequality
an+1 + bn+1 + cn+1
a+b+c
≥
.
n
n
n
a +b +c
3
(Hint: Chebyshev’s inequality.)
12. Solve the problem from Example 21 by using Chebyshev’s inequality.
(Hint: Write ln(aa bb cc ) as a ln a + b ln b + c ln c.)
13. (IMO, 1978) Let a1 , a2 , ..., an be distinct positive integers. Prove that
a1 a2
an
1 1
1
+ 2 + ··· + 2 ≥ + + ··· + .
2
1
2
n
1 2
n
(Hint: The Rearrangement inequality.)
14. Let x1 , x2 , ..., xn be positive numbers. Show that
(Hint: The Rearrangement inequality.)
19
x1 x2
xn
+
+ ··· +
≥ n.
x2 x3
x1
15. Let a, b, c be positive real numbers. Use the Rearrangement inequality in order to prove that
a8 + b 8 + c 8
1 1 1
≥ + + . (Compare with Example 28 above).
3
3
3
abc
a b c
16. (Korea, 2000) The real numbers a, b, c, x, y, z satisfy a ≥ b ≥ c > 0 and x ≥ y ≥ z > 0
b2 y 2
c2 z 2
3
a2 x 2
+
+
≥ .
Prove that
(by + cz)(bz + cy) (cz + ax)(cx + az) (ax + by)(ay + bx)
4
(Hint: With the notation α = a2 x2 , β = b2 y 2 , γ = c2 z 2 and using the Rearrangement inequa1
α
β
γ lity, show that the expression is ≥
+
+
. Now use the Cauchy-Schwarz
2 β+γ
γ+α
α+β
inequality.)
17. (Ireland, 1999) Let a, b, c, d be positive real numbers whose sum is 1. Prove that
a2
+
a+b
b2
c2
d2
1
+
+
≥ with equality if and only if a = b = c = d = 1/4.
b+c c+d d+a
2
(Hint: Apply the Cauchy-Schwarz inequality to the left-hand side multiplied by (a + b) + (b +
c) + (c + d) + (d + a).)
18. (Czech Republic and Slovakia, 1999) For arbitrary positive real numbers a, b, c, prove the
a
b
c
+
+
≥ 1.
inequality
b + 2c c + 2a a + 2b
(Hint: Apply the Cauchy-Schwarz inequality to the left-hand side multiplied by a(b + 2c) +
b(c + 2a) + c(a + 2b) or substitute x = b + 2c, y = c + 2a, z = a + 2b and use AM-GM.)
a3 b 3
19. Let a, b, c, d be positive real numbers such that c +d = (a +b ) . Show that + ≥ 1.
c
d
(Hint: Using the Cauchy-Schwarz inequality show that (a3 /c + b3 /d)(ac + bd) ≥ (a2 + b2 )2 ≥
ac + bd.)
2
2
2
2 3
20. For non negative real numbers x√1 , x2 , ..., xn show
n that
(1 + x1 )(1 + x2 ) · · · (1 + xn ) ≥ 1 + n x1 x2 · · · xn .
√
(Hint: Let ai = n xi and then apply Hölder’s inequality.)
21. Use Hölder’s inequality to prove that a3 + b3 + c3 ≥ a2 b + b2 c + c2 a for all positive real
numbers a, b, c.
22. Given a, b, c, p, q, r, positive real numbers with p + q + r = 1, prove that a + b + c ≥
a b c + aq b r c p + ar b p c q .
(Hint: Hölder’s inequality.)
p q r
n
n
X
X
1 2
23. Let xk ∈ [1, 2], k = 1, 2, ..., n. Prove that 2
xk ·
≥ n3 .
xk
k=1
k=1
(Hint: Hölder’s inequality with p = 13 , q = 23 .)
20
r
24. Let a, b, c, x, y, x be positive real numbers. Prove that
r
r
ay 2 + bz 2 + cx2
az 2 + bx2 + cy 2
+
+
≥ x + y + z.
a+b+c
a+b+c
(Hint: Minkowski’s inequality.)
ax2 + by 2 + cz 2
a+b+c
1 1 4 16
64
+ + +
≥
.
a b c
d
a+b+c+d
(Observe the improvement compared with the problem 3 above. Hint: Hölder’s inequality.)
25. Let a, b, c, d be positive real numbers. Prove that
26. (Ireland, 1998) Let a, b, c be positive real numbers. Apply Jensen’s inequality to show that
9
1
1
1 1 1 1
≤2
+
+
≤ + + .
a+b+c
a+b b+c c+a
a b c
x
+
27. Let x, y, z be three positive real numbers such that x + y + z = 1. Prove that
x+1
y
z
3
+
≤ .
y+1 z+1
4
(Hint: Jensen’s or the AM-GM inequality).
28. Find the minimum value of the expression
0 < a1 , a2 , ..., an < 2 and
n
X
n
X
k=1
1
, where
4ak − a3k
ak = n.
k=1
(Hint: Jensen’s inequality or the AM-GM and Hölder’s inequalities.)
29. (Nordic, 1992) Show that among all triangles with the inscribed circle of radius 1, the
equilateral triangle has the minimum circumference.
(Hint: Letting 2α, 2β, 2γ be angles of the triangle, half the circumference may be expressed as
1
f (α) + f (β) + f (γ), where f (x) =
, for 0 < x < π/2.)
tan x
x, y, z be three
r 30. Let r
r real numbers such that 0 < x, y, z < 4 and xyz = 1. Prove that
x
y
z
+
+
≥ 1.
x+8
y+8
z+8
(Hint: Looks like a candidate for Jensen’s inequality. However, in this problem we have a
product condition xyz = 1, where we instead would like to have the sum condition x + y +
z = constant. This obstacle can however be removed by introducing new variables a, b, c by
es 1/2
substitutions x = ea , y = eb and z = ec . Now, study the function f (s) = s
. This is in
e +8
fact a part of a problem from IMO 2001. For the whole problem you must remove the constraint
x, y, z < 4. If only one of the variables is ≥ 4 you can still use Jensen’s inequality (for those two
terms with variable values less than 4). If at least two of variables are greater than 4 the proof is
straightforward.)
31. (Russia, 1999) Let x and y bet two positive real numbers such that x3 + y 3 > 2. Prove that
21
x2 + y 3 < x 3 + y 4 .
(Hint: You may start with the Power Mean inequality for (x2 + y 2 )/2 and (x3 + y 3 )/2.
32. (Iran, 1998) Let a1 , a2 , a3 , a4 be positive real numbers such that
4
4
4
nX
X
X
1o
a1 a2 a3 a4 = 1. Prove that
a3i ≥ max
ai ,
.
a
i
i=1
i=1
i=1
(Hint: For the first inequality you may use the Power Mean inequality. The second follows from
the AM-GM inequality.)
33. Let a, b, c, d be positive real numbers. Prove that
a3 + b3 + c3 a3 + b3 + d3
+
+
a+b+c
a+b+d
a3 + c3 + d3 b3 + c3 + d3
+
≥ a2 + b2 + c2 + d2 .
a+c+d
b+c+d
1
(Hint: Using the Power Mean inequality show that a3 + b3 + c3 ≥ (a + b + c)(a2 + b2 + c2 ).)
3
34. (Poland, 1999)pLet x, y, z be three positive real numbers such that x + y + z = 1. Prove
that x2 + y 2 + z 2 + 2 3xyz ≤ 1.
(Hint: Make the inequality homogeneous.)
35. (UK, 1999) Some three nonnegative real numbers p, q, r satisfy p + q + r = 1. Prove that
7(pq + qr + rp) ≤ 2 + 9pqr.
(Hint: Use the condition p + q + r = 1 to make the inequality homogeneous and then use
Schur’s inequality.)
36. Let a1 , a2 , ..., an be positive real numbers such that (1 + a1 )(1 + a2 ) · · · (1 + an ) = 2n .
Prove that a1 a2 · · · an ≤ 1.
(Hint: With the notation of the Maclaurin’s inequality write (1 + a1 )(1 + a2 ) · · · (1 + an ) as a
sum of S1 , ..., Sn and then show that this sum is ≥ (1 + Sn1/n )n . The binomial theorem may be
useful here. Or you may use the same metod as in problem 20.)
a
37. (Asian-Pacific MO, 1998) Let a, b, c be positive real numbers. Prove that 1 +
1+
b
c
a + b + c
b
1+
≥2 1+ √
.
3
c
a
abc
(Hint: Make the substitution a = x3 and so on, then apply Muirhead’s inequality.)
x
y
z
9
+
+
≤
(x + y)(x + z) (y + z)(y + x) (z + x)(z + y)
4(x + y + z)
for all positive real numbers x, y, z.
(Hint: Use the Muirhead’s inequality.)
38. Prove the inequality
39. p
For real numbers
√ −1 ≤px, y ≤ 1, find the maximum value of the function f (x, y) =
2
xy + x 1 − y + y 1 − x2 − (1 − x2 )(1 − y 2 ).
(Hint: The expression invites to the trigonometric substitution x = cos α, y = cos β, for 0 ≤
α, β ≤ π.)
22
40. Let x, y, z be positive real numbers such that xy + yz + zx = 1. Prove the following
x
y
z
2x(1 − x2 ) 2y(1 − y 2 ) 2z(1 − z 2 )
+
+
≤
+
+
.
inequality
2
2
2
2
2
2
2
2
(1 + x )
(1 + y )
(1 + z )
1+x
1+y
1 + z2
2 tan α2
1 − tan2 α2
(Hint: Since the expressions are similar to those of sin α =
and
cos
α
=
,
1 + tan2 α2
1 + tan2 α2
you can try to use the substitution x = tan α2 , y = tan β2 , z = tan γ2 .)
PROOFS OF THE INEQUALITIES
1. AM-GM-HM inequality
Considering the importance and usefulness of the AM-GM inequality, we give here three different
proofs of it. Yet another proof can be found in the last section of this text.
Proof 1 (by induction): We start by proving the inequality for n = 2.
√
√
We have ( ai − a2 )2 ≥ 0. We develop the left-hand side of this inequality, move the negative
a1 + a2 √
term over to the right-hand side and then divide both sides by 2 to get
≥ a1 a2 . Thus the
2
AM-GM inequality holds for n = 2.
Now assume that the inequality holds for some n ≥ 2. Then we will show that it holds for 2n.
We have
2n
X
i=1
vv
uu
u uY
n
ut
n
2nt
ai ·
i=1
ai =
n
X
i=1
ai +
v
n
u
uY
n
t
ai +
ai ≥ n
2n
X
i=n+1
i=1
v
u 2n
u Y n
t
ai ≥
i=n+1
v
v
u 2n
u 2n
uY
u Y
n
t a.
t
ai = 2n 2n
i
i=n+1
i=1
The first inequality in the expression above holds according to our assumption. The second
inequality follows from the AM-GM inequality for n = 2.
Thus, by the induction argument the AM-GM inequality holds for n = 2k , (k = 1, 2, ...).
At last we assume that the inequality holds for some n ≥ 2. Then we will show that it holds for
n − 1.
v
un−1
uY
n−1
ai . Then we have
Let a1 , a2 , ..., an−1 be positive real numbers and let an = g = t
i=1
23
n−1
X
v
un−1
uY
p
n
ai + g ≥ n t
ai · g = n n g n−1 · g = ng. Hence we have
i=1
i=1
v
v
v
un−1
u
un−1
n−1
n−1
n−1
uY
uY
uY
X
X
n−1
n−1
n−1
t
t
ai ≥ n
ai , i.e.
ai .
ai +
ai ≥ (n − 1) t
i=1
i=1
i=1
i=1
i=1
The inequality in the expression above holds according to our assumption.
Thus the AM-GM inequality holds for all n ≥ 2.
Proof 2 (by easy analysis):
First we prove one useful theorem, which in fact is a generalization of the AM-GM inequality
(the weighted version).
Theorem 1: Let a1 , a2 , ..., an be positive real numbers. Let then α1 , α2 , ..., αn be positive real
n
n
Y
X
αk
numbers such that α1 + α2 + ... + αn = 1. We set G =
ak and A =
αk ak . Then we have
G ≤ A with equality if and only if a1 = a2 = ... = an .
k=1
k=1
Proof: We set ak = (1 + xk )A. We note that xk > −1, (k = 1, 2, ..., n) and that
Hence G =
n
Y
k=1
aαk k =
n
Y
((1 + xk )A)αk = A
k=1
One realizes that the inequality A
n
Y
n
Y
(1 + xk )αk ≤ A
k=1
(1 + xk )αk ≤ A
k=1
n
Y
n
Y
n
X
αk xk = 0.
k=1
exk αk = A.
k=1
exk αk holds by studying the function
k=1
f (x) = ex − (1 + x). (We have f (0) = 0, and since f 0 (x) = ex − 1, we have f 0 (x) = 0 for x = 0.
Finally we have f 00 (0) = 1, which means that f (x) has minimum at x = 0. Thus, f (x) ≥ 0 for all
x ∈ R). We get equality only when xk = 0, (k = 1, 2, ..., n), i.e. when a1 = a2 = ... = an .
1
Now, if we in Theorem 1 set αk = , (k = 1, 2, ..., n), we get
n
√
a1 + a2 + ... + an
n
, which is the AM-GM inequality.
a1 · a2 · ... + ·an ≤
n
Proof 3: The AM-GM inequality also follows easily from Jensen’s inequality.
Let a1 , a2 , ..., an be positive real numbers. We note that the function f (x) = ex is strictly
1
convex, and that n · = 1. Jensen’s inequality (with xi = ln ai , i = 1, 2, ..., n) then yields
n
1
1
1
1
1
1
e( n ln a1 + n ln a2 +...+ n ln an ) ≤ eln a1 + eln a2 + ... + eln an ,
n
n
n
which after simplifying becomes the AM-GM inequality.
24
It may also be worth pointing out that the AM-GM inequality follows directly from the Power
Mean inequality. (Just keep the first and the last part of that inequality and you get the AM-GM
inequality.)
Proof of the GM-HM inequality: Follows easily from the AM-GM inequality:
1
1 1
are also positive real numbers.
Let a1 , a2 , ..., an be positive real numbers. Then , , ...,
an
r a1 a2
1
+ a12 + ... + a1n
1 1
1
a1
n
According to the AM-GM inequality, we have
·
· ... ·
≤
. When we
a1 a2
an
n
invert both sides of the inequality, then, of course, the sign of the inequality gets reversed. Hence
√
n
n
a1 · a2 · ... · an ≥ 1
1
1 .
+
+
...
+
a1
a2
an
2. Chebyshev’s inequality
We note that
n X
n
X
(ai bi − ai bj ) =
i=1 j=1
n
n
n
X
ai b i −
n
X
ai
n
X
i=1
i=1
i=1
n
X
n
X
n
X
j=1
aj b j −
j=1
aj
n
X
(nai bi − ai
i=1
bi , and that
bi = n
n
X
i=1
i=1
n
X
n
X
bj ) = n
j=1
n X
n
X
i=1 j=1
n
X
ai b i −
(aj bj − aj bi ) =
ai
i=1
n
X
n
X
n
X
i=1
ai
i=1
n X
n
X
i=1
n
X
ai b i −
n
X
n
X
bj =
j=1
aj b j −
j=1
n
X
aj · b i =
j=1
bi .
i=1
n
n
n
n
1 XX
1 XX
It follows that n
ai b i −
ai
bi =
(ai bi − ai bj ) +
(aj bj − aj bi ) =
2 i=1 j=1
2 i=1 j=1
i=1
i=1
i=1
n
n
n
n
1 XX
1 XX
(ai bi − ai bj + aj bj − aj bi ) =
((ai − aj )(bi − bj )).
2 i=1 j=1
2 i=1 j=1
Now we realize that (ai − aj )(bi − bj ) ≥ 0 for i, j = 1, 2, ..., n. (The conditions
ai ≤ a2 ≤ ... ≤ an , bi ≤ b2 ≤ ... ≤ bn lead to the fact that the two factors on the left-hand side of
n
n
n
X
X
X
the last mentioned inequality will always have the same sign.) Hence, n
ai b i −
ai
bi ≥ 0.
i=1
i=1
i=1
Moving the right term on the left-hand side of this inequality to the right-hand side and then
dividing both sides by n2 yields Chebyshev’s inequality.
It is not hard to realize that in Chebyshev’s inequality we will have equality if and only if
a1 = a2 = ... = an or b1 = b2 = ... = bn This follows from the facts that (ai − aj )(bi − bj ) ≥ 0,
for i, j = 1, 2, ..., n. To realize that the inequality is reversed if we have ai ≤ a2 ≤ ... ≤ an and
bi ≥ b2 ≥ ... ≥ bn is no more difficult. This follows from (ai −aj )(bi −bj ) ≤ 0, for i, j = 1, 2, ..., n
(these inequalities are reversed because now the conditions lead to the fact that the two factors on
25
the left-hand side will always have the opposite sign).
Another proof: We will now show that Chebyshev’s inequality easily can be derived from the
Rearrangement inequality. Chebyshev’s inequality can be expressed as (a1 + a2 + ... + an )(b1 +
b2 + ... + bn ) ≤ n(a1 b1 + a2 b2 + .. + an bn ). On the left-hand side we multiply the two parenthesis
together. We then get n2 terms on that side. Then we arrange these terms in n groups as follows:
(a1 b1 + a2 b2 + ... + an bn ) + (a1 b2 + a2 b3 + ... + an b1 ) + ... + (a1 bn + a2 b1 + ... + an bn−1 ). According
to the Rearrangement inequality a1 b1 + a2 b2 + .. + an bn ≥ anyone of these n groups of numbers,
and Chebyshev’s inequality follows. Using the same technique one can easily show the case when
Chebyshev’s inequality is reversed.
3. Rearrangement inequality
We start by showing the inequality for n = 2. Let a1 ≤ a2 and bi ≤ b2 be real numbers. Then
we have (a2 − a1 )(b2 − b1 ) ≥ 0, since both factors on the left-hand side of the inequality are
definitely nonnegative. Multiplying these factors together and then rearranging the terms yields
the inequality a1 b1 + a2 b2 ≥ a1 b2 + a2 b1 , which is the Rearrangement inequality for n = 2. We
realize that we have equality only if a1 = a2 or b1 = b2 .
Now we turn to the general case.
Let b1 ≤ b2 ≤ ... ≤ bn and c1 , c2 , ..., cn be real numbers. Let then a1 , a2 , ..., an be a permutation
of c1 , c2 , ..., cn such that a1 b1 + a2 b2 + ... + an bn is maximal. Now, assume that we for some i < j
have ai > aj . Then we have ai bj +aj bi ≥ ai bi +aj bj (the case n = 2). Thus, a1 b1 +a2 b2 +...+an bn
is not maximal unless a1 ≤ a2 ≤ ... ≤ an or bi = bj for all i < j such that ai > aj . In the
later case the numbers ai , aj can change places so that we get a1 ≤ a2 ≤ ... ≤ an . Hence,
a1 b1 + a2 b2 + ... + an bn is maximal when a1 ≤ a2 ≤ ... ≤ an .
Now we note that −(a1 b1 + a2 b2 + ... + an bn ) = (−a1 )b1 + (−a2 )b2 + ... + (−an )bn is
maximal when −a1 ≤ −a2 ≤ ... ≤ −an and b1 ≤ b2 ≤ ... ≤ bn . From that follows that
a1 b1 + a2 b2 + ... + an bn is minimal under the same conditions. But the condition −a1 ≤ −a2 ≤
... ≤ −an is equivalent to the condition a1 ≥ a2 ≥ ... ≥ an . Thus, a1 b1 + a2 b2 + ... + an bn is
minimal when b1 ≤ b2 ≤ ... ≤ bn and a1 ≥ a2 ≥ ... ≥ an .
4. Cauchy-Schwarz inequality
Let a1 , a2 , ..., an and b1 , b2 , ..., bn be real numbers. Let us look at the polynomial
n
X
i=1
i=1
n
n
X
X
2
2
ai x + 2
ai b i x +
b2i .
i=1
n
X
i=1
26
(ai x + bi )2 =
Since
n
X
n
X
2
(ai x + bi ) ≥ 0 for all real numbers x, it is obvious that
n
X
i=1
a2i
n
X
x +2
ai b i x +
2
i=1
i=1
b2i ≥ 0 too. It follows that the discriminant for this polynomial is ≤ 0. The discriminant for a
i=1
n
n
n
X
2
X
X
quadratic polynomial ax2 + bx + c is b2 − 4ac. Thus, 4
ai b i − 4
b2i ≤ 0. If we
a2i
i=1
i=1
i=1
then move the right-hand term on the left-hand side of this inequality to the right-hand side and
finally divide both sides by 4, we end up with the Cauchy-Schwarz inequality.
So, why does the discriminant for the above polynomial have to be ≤ 0? If we solve the
equation ax2 + bx + c = 0 for arbitrary real numbers a, b, c using quadratic completion, we get
1√ 2
b
the solution x = − ±
b − 4ac. Now, since we have ax2 + bx + c ≥ 0, we know that the
2a 2a
above equation will have no real roots or one double root. If we have no real roots, the sum of the
terms under the root sign will be negative, i.e. b2 − 4ac < 0. If we have a double root, we will have
b2 − 4ac = 0.
It follows from the fact that
n
X
(ai x + bi )2 ≥ 0 for all real numbers x, that the Cauchy-Schwarz
i=1
inequality will have equality if and only if the number sequences a and b are proportional.
Another proof of this inequality is supplied in the last part of this text.
5. Hölder’s inequality
n
n
n
1t
X
1s X
X
1
1
btk .
and q = . Then Hölder’s inequality becomes
ak b k ≤
ask ·
s
t
k=1
k=1
k=1
n X
ask 1s btk 1t
Pn s
The inequality can then be expressed in the following way:
· Pn t
≤ 1.
a
b
i
i
i=1
i=1
k=1
1 1
We note that + = 1. Then, applying Theorem 1 (see proof of the AM-GM inequality) to the
s
t
left-hand side of this inequality yields
n n X
X
ask 1s btk 1t
1 ask
1 btk 1 1
Pn s
P
· Pn t
≤
+ Pn t = + = 1.
s ni=1 asi
t i=1 bi
s t
i=1 ai
i=1 bi
k=1
k=1
Set p =
We have equality if and only if the number sequences (as1 , as2 , ..., asn ) and (bt1 , bt2 , ..., btn ) are
proportional. This follows from the conditions for equality in Theorem 1.
We will now show an additional useful result:
If we in Hölder’s inequality allow one of the parameters p, q to be negative, then the inequality is
reversed.
27
s
1
Suppose that p < 0, i.e. s < 0 with the notation above. We set S = − and T = . Then we
t
t
1
1
−t
t t
have S > 0, T > 0 and + = 1. Now we set uk = ak and vk = ak bk , (k = 1, 2, ..., n). Then
S nT
n
n
X
1 X
T1
X
S S
Hölder’s inequality yields
uk v k ≤
·
uk
vkT , which of course is equivalent to
k=1
n
X
t t
a−t
k ak b k ≤
n
X
k=1
k=1
k=1
n
t X
t
−t·(− st ) − s
t t 1t
·
(ak bk ) . After some simplifying, rearranging and then
ak
k=1
k=1
taking the t:th root of both sides of the inequality, we get
n
X
ak b k ≥
k=1
n
X
ask
n
1
1s X
t t
·
bk , which
k=1
k=1
is the reversed inequality.
6. Minkowski’s inequality
We note that
n
X
r
(ai + bi ) =
i=1
n
X
r−1
(ai + bi )(ai + bi )
=
n
X
i=1
r−1
ai (ai + bi )
i=1
+
n
X
bi (ai + bi )r−1 .
i=1
1 1
r
For r > 1 we set s so that + = 1, i.e. s =
. Then, by Hölder’s inequality, we have the
r
s
r−1
n
n
X
X
r−1
following (we apply the inequality to both of the terms
ai (ai + bi )
and
bi (ai + bi )r−1 ):
n
X
(ai + bi )r ≤
i=1
n
X
i=1
n
X
i=1
i=1
n
n
n
1 X
1 X
1 X
1
r r
(r−1)·s s
r r
(r−1)·s s
ai
·
+
·
=
(ai + bi )
bi
(ai + bi )
i=1
i=1
i=1
i=1
n
n
n
X
r−1
1 X
r−1
1 X
r
r
r
r r
r
r r
(ai + bi )
+
bi ·
(ai + bi )
. If we then divide both sides of
ai ·
i=1
this inequality by
n
X
i=1
(ai + bi )r
r−1
r
i=1
, we get Minkowski’s inequality.
i=1
It is obvious that Minkowski’s inequality holds for r = 1; that condition yields equality. When
r > 1 we have equality if and only if the number sequences a and b are proportional. This follows
from the conditions for equality in Hölder’s inequality.
If r < 1, r 6= 0, then s becomes negative, which leads to Minkowski’s inequality getting
reversed. This follows from the fact that Hölder’s inequality becomes reversed for s < 0.
7. Jensen’s inequality
We will prove Jensen’s inequality for rational positive numbers α1 , α2 , ..., αn only. The general
28
proof for nonnegative real numbers α1 , α2 , ..., αn requires some serious analytical arguments.
We will divide the proof of into two cases.
1
Case 1: αi = , (i = 1, 2, ..., n). (We use the same approach as in the first proof of the AM-GM
n
inequality.)
In this case we are supposed to prove the inequality
n
n
1 X
1X
(1) f
xi ≤
f (xi ). It is true for n = 2. This follows directly from the definition
n i=1
n i=1
of convexity. Now, assume that (1) holds for n = 2k , (k = 1, 2, ...). Then (1) also holds for
m = 2k+1 = 2n.
Proof: Let x1 , x2 , ..., xm ∈ I. Then we have
P
P
n
n
P
P
1
1
n
n
1
1
x + x + ... + x xi + n i=1 xn+i f n i=1 xi + f n i=1 xn+i
1
2
m
= f n i=1
≤
≤
f
m
2
2
Pn
Pn
Pm
f (xi )
i=1 f (xn+i )
i=1 f (xi ) +
= i=1
.
2n
m
In the expression above, the first inequality holds because (1) holds for n = 2. (We have that
n
n
1X
1X
the two numbers
xi ,
xn+i ∈ I, since they are the arithmetic mean of n numbers that
n i=1
n i=1
all ∈ I.) The second inequality follows from our assumption. Thus, since (1) holds for n = 2, by
induction it holds for n = 2k , (k = 1, 2, ...).
Assume that (1) holds for n > 2. Then (1) also holds for n − 1
1
(x1 +
n−1
x2 + ... + xn−1 ) (the arithmetic mean of x1 , x2 , ..., xn−1 ), (1) holds according to our assumption.
We get
Proof: Let x1 , x2 , ..., xn−1 ∈ I. For these numbers together with the number xn =
(2)
f
x1 + x2 + ... + xn−1 +
x1 +x2 +...+xn−1
n−1
n
f (x1 ) + f (x2 ) + ... + f (xn−1 ) + f
x1 +x2 +...+xn−1
n−1
≤
. After simplifying, the left-hand side of (2)
n
n−1
x + x + ... + x x + x + ... + x 1 X
1
2
n−1
1
2
n−1
becomes f
. Thus, f
≤
f (xi )+
n−1
n−1
n i=1
x + x + ... + x 1 x1 + x2 + ... + xn−1 1
2
n−1
f
. Further simplifying yields the inequality f
≤
n
n−1
n−1
n−1
1 X
f (xi ). Now, by induction, (1) holds for n ≥ 2. Hence, (1) is proved in case 1.
n − 1 i=1
29
Case 2: α1 , α2 , ..., αn are rational positive numbers.
Since α1 , α2 , ..., αn are rational positive numbers, there is a natural number m and nonnegative
pi
integers p1 , p2 , ..., pn such that m = p1 + p2 + ... + pn and αi = , (i = 1, 2, ..., n). (To realize
m
this, just rewrite the numbers α1 , α2 , ..., αn so that they all have the same denominator.)
Case 1 yields
(x + ... + x ) + ... + (x + ... + x ) 1
1
n
n
≤
(3) f
m
(f (x1 ) + ... + f (x1 )) + ... + (f (xn ) + ... + f (xn ))
. (The first parenthesis in the nominator on
m
the left-hand side of the inequality contains p1 terms, the second parenthesis p2 terms, and so on.)
n
n
1 X
1 X
Now, (3) can be expressed as f
pi xi ≤
pi f (xi ). Hence, (1) is proved in case 2. m i=1
m i=1
8. Power Mean inequality
We start by pointing out the obvious: for k = m we have equality, which means that the inequality
m
> 1 and it follows that
holds in that case. Now we assume instead that k < m. Then we have
k
m
mm
00
k
the function f (x) = x is strictly convex for x ≥ 0. (The second derivative, f (x) =
−
k k
m
1 x k −2 > 0 for x > 0). Now, since the numbers a1 , a2 , ..., an are nonnegative, clearly the numbers
ak1 , ak2 , ..., akn are nonnegative too. With help from Jensen’s inequality we then get the inequality
1 k m 1 k 1 k
1 k m 1 k m
1 k mk
k
k
k
(a ) + (a2 ) + ... + (an ) ≥
a + a + ... + an , which, after some simplifying,
n 1
n
n
n 1 n 2
n m
1 m
1 k
k
m
k
m
k
becomes (a1 + a2 + ... + an ) ≥
(a1 + a2 + ... + an ) . Taking the m:th root of both sides
n
n
of this inequality then yields the Power Mean inequality.
The condition for equality follows from the condition for equality in Jensen’s inequality.
9. Schur’s inequality
We will derive Schur’s inequality from a stronger theorem.
Theorem 2: If a, b, c, u, v, w are nonnegative real numbers and we have
1
1
1
(1) a p + c p ≤ b p and
1
1
1
(2) u p+1 + w p+1 ≥ v p+1 then, if p > 0, we have
(3) ubc − vca + wab ≥ 0.
1
1
Proof: We start with two pairs of nonnegative numbers: (a p+1 , c p+1 ) and
30
1
1
((uc) p+1 , (wa) p+1 ). Since we have p > 0, we have
1
1
p
p
1
1
> 0,
> 0 and
+
= 1.
p+1
p+1
p+1 p+1
1
1
Then Hölder’s inequality yields a p+1 (uc) p+1 + c p+1 (wa) p+1 ≤
1
≤ (a p+1 ·
1
p+1
p
1
1
+ c p+1 ·
p+1
p
1
p
1
1
1
) p+1 · ((uc) p+1 ·(p+1) + (wa) p+1 ·(p+1) ) p+1 , i.e.
1
1
1
p
1
(ac) p+1 u p+1 + (ac) p+1 w p+1 ≤ (a p + c p ) p+1 · (uc + wa) p+1 . Taking the (p + 1):th power of both
1
1
1
1
sides of this inequality, we get ac(u p+1 + w p+1 )p+1 ≤ (a p + c p )p · (uc + wa). Now we use the
1
1
conditions (1) and (2) in order to get ac(v p+1 )p+1 ≤ (b p )p (uc + wa), which is equivalent to
(3) ubc − vca + wab ≥ 0.
Now we can assume that 0 ≤ z ≤ y ≤ x. Then, using Theorem 2 (setting p = 1, a = y − z,
b = x − z, c = x − y, u = xr , v = y r , w = z r ), we get xr (x − z)(x − y) − y r (x − y)(y − z) +
z r (y − z)(x − z) ≥ 0, which is equivalent to Schur’s inequality.
It is quite easy to show that we have equality if and only if and only if x = y = z or if two of
x, y, z are equal and the third is 0. The only one of the three terms on the left-hand side of Schur’s
inequality that can be negative is y r (y − x)(y − z). It is negative when we have y < x and y > z.
But with those conditions we see that xr (x − z)(x − y) > |y r (y − x)(y − z)|. Thus, to have equality
we must have y = x or y = z. In either of these two cases two of the three terms on the left-hand
side of the inequality = 0. It follows that in order to have equality we must have x = y = z or two
of x, y, z must be equal and the third must be 0.
10. Maclaurin’s inequality
We start by proving a theorem that we need in order to establish MacLaurin’s inequality.
Theorem 3: For n ≥ 2, let a1 , a2 , ..., an be positive real numbers that are not all equal. Also let
X
ai1 ai2 ...air
p0 = 1 and pr =
, r = 1, 2, ..., n.
n
1≤i1<i2<...<ir≤n
r
Then, for all r, 1 ≤ r < n, we have pr−1 pr+1 < p2r .
a1 + a2
Proof (by induction): For n = 2 we have p0 = 1, p1 =
and p2 = a1 a2 . Thus,
2
a + a 2
1
2
p0 p2 = a1 a2 <
= p21 . Of course the inequality above follows from the AM-GM
2
inequality (since we have a1 6= a2 , we have a strict inequality). Hence we have proved the theorem
for n = 2.
Suppose now that the theorem holds for some n = k − 1, k ≥ 3. We will show that it then
holds for n = k.
According to our assumption, for the positive real numbers a1 , a2 , ..., ak−1 that are not all equal
we have Pr−1 Pr+1 < Pr2 , 1 ≤ r < k − 1. (Here we write P instead of p to be able to separate
these numbers that form the inequality for n = k − 1 from the numbers that form the inequality
31
for n = k; the inequality that we are supposed to prove.) We want to show that for the positive real
numbers a1 , a2 , ..., ak that are not all equal, the inequality pr−1 pr+1 < p2r , 1 ≤ r < k, holds.
k−r
r
Pr + ak Pr−1 , r = 1, 2, ..., k, where we set Pk = 0. (This can be
k
k
n
n−1
n−1
a little tricky to realize, but it follows from the fact that
=
+
.) We get
k
k
k−1
k − r − 1
k − r + 1
r−1
r+1
Pr−1 +
ak Pr−2
Pr+1 +
ak P r −
(1) k 2 (pr−1 pr+1 −p2r ) = k 2
k
k
k
k
k − r
2
r
Pr + ak Pr−1
= A + Bak + Ca2k , where
k
k
Now observe that pr =
A = ((k − r)2 − 1)Pr−1 Pr+1 − (k − r)2 Pr2 ,
B = (k − r + 1)(r + 1)Pr−1 Pr + (k − r − 1)(r − 1)Pr−2 Pr+1 − 2(k − r)rPr Pr−1 ,
2
C = (r2 − 1)Pr−2 Pr − r2 Pr−1
.
Since not all a1 , a2 , ..., ak−1 are equal, we have Pr−1 Pr+1 < Pr2 ,
2
Pr−2 Pr < Pr−1
and Pr−2 Pr+1 < Pr−1 Pr . (The last inequality is valid because Pr−2 Pr+1 =
Pr−1 Pr−1 Pr+1
Pr−1 Pr2
Pr+1
2 Pr+1
< Pr−1
=
<
= Pr−1 Pr .)
Pr−2 Pr
Pr
Pr
Pr
Pr
2
. Furthermore,
From these inequalities follow that A < −Pr2 , B < 2Pr−1 Pr and C < −Pr−1
2
2
2
2
2
from (1) we get k (pr−1 pr+1 − pr ) < −Pr + 2Pr Pr−1 ak − Pr−1 ak = −(Pr − Pr−1 ak )2 ≤ 0. Thus,
k 2 (pr−1 pr+1 − p2r ) < 0, which is equivalent to pr−1 pr+1 < p2r . Hence the theorem is proved for all
n ≥ 2.
One more thing remains, though; the case where a1 = a2 = ... = ak−1 6= ak . In that case we
Pr
2
2
2
have a1 =
, and from (1) we get k 2 (pr−1 pr+1 − p2r ) = −a21 Pr−1
+ 2a1 Pr−1
ak − Pr−1
a2k =
Pr−1
2
−(a1 − ak )Pr−1
< 0. Thus the theorem is proved in that case also.
Proof of MacLaurin’s inequality: Consider the inequalitites
p0 p2 < p21
(p1 p3 )2 < p42
(p2 p4 )3 < p63
.
.
.
(pr−1 pr+1 )r < p2r
r
If we multiply together all the left-hand and right-hand sides respectively, we come up with
r−1 r
2r
r
r+1
the inequality p0 p21 p42 p63 ...p2r−2
pr+1 < p21 p42 p63 ...p2r−2
r−1 pr √
r−1 pr , which yields p
r+1 < p√
r . The last
√
√
r
3
r+1
pr+1 . Hence, we have p1 > p2 > p3 > ... >
inequality can also be written as pr >
32
√
n−1
pn−1 >
√
n
pn , which is MacLaurin’s inequality.
In the case where a1 = a2 = ... = an , we have p1 =
√
n
pn . This is easy to show. In that case we have
s
s
n
1
1 n k √
k
k
a1 =
a1 = k p k .
n
n
k
k
k
k
√
k−1
√
pk−1 =
p2s=
k−1
√
3
√
p3 = ... = n−1 pn−1 =
1
n
a1k−1 = a1 =
n
k−1
k−1
11. Muirhead’s inequality
We start by proving the following lemma:
Lemma: Let a1 , a2 , b1 , b2 be nonnegative real numbers such that a1 + a2 = b1 + b2 and
max(a1 , a2 )≥max(b1 , b2 ). Let also x, y be nonnegative real numbers. Then we have
(0)
xa1 y a2 + xa2 y a1 ≥ xb1 y b2 + xb2 y b1 .
Proof: Because of symmetry we can without loss of generality assume that a1 ≥ a2 and b1 ≥
b2 . If any of the numbers x, y = 0 then (0) obviously holds. Therefore we assume that x, y 6= 0. We
have xa1 y a2 + xa2 y a1 − xb1 y b2 − xb2 y b1 = xa2 y a2 (xa1 −a2 + y a1 −a2 − xb1 −a2 y b2 −a2 − xb2 −a2 y b1 −a2 ) =
xa2 y a2 (xb1 −a2 − y b1 −a2 )(xb2 −a2 − y b2 −a2 ) ≥ 0.
Now, why does xa2 y a2 (xb1 −a2 − y b1 −a2 )(xb2 −a2 − y b2 −a2 ) ≥ 0 hold? Clearly, we have xa2 y a2 ≥
0. Then, from the conditions the inequalities b1 − a2 ≥ 0 and b2 − a2 ≥ 0 follow. If we then have
x ≥ y, we have xb1 −a2 − y b1 −a2 ≥ 0 and xb2 −a2 − y b2 −a2 ≥ 0, and the inequality follows. If we on
the contrary have x ≤ y, we have xb1 −a2 − y b1 −a2 ≤ 0 and xb2 −a2 − y b2 −a2 ≤ 0, and the inequality
follows in that case too.
We divide the proof of Muirhead’s inequality into two cases.
Case 1: b1 ≥ a2 .
According to the conditions, we have a1 ≥ a1 + a2 − b1 and a1 ≥ b1 . Thus, max(a1 , a2 )≥
max(a1 + a2 − b1 , b1 ). We also note that a1 + a2 = (a1 + a2 − b1 ) + b1 . We use the lemma
and get xa1 y a2 + xa2 y a1 ≥ xa1 +a2 −b1 y b1 + xb1 y a1 +a2 −b1 . Clearly also z a3 (xa1 y a2 + xa2 y a1 ) ≥
z a3 (xa1 +a2 −b1 y b1 + xb1 y a1 +a2 −b1 ) holds. Then we realize that
X
X
(1)
z a3 (xa1 y a2 + xa2 y a1 ) ≥
z a3 (xa1 +a2 −b1 y b1 + xb1 y a1 +a2 −b1 ) holds too. (When we
cyclic
cyclic
used the lemma, we could for example have used the numbers y and z instead of the numbers x
and y and then multiplied both sides of the inequality with xa3 instead of z a3 .)
X
X
Now, we note that
z a3 (xa1 y a2 + xa2 y a1 ) =
x a1 y a2 z a3 .
sym
cyclic
According to the conditions, we also have a1 + a2 − b1 ≥ b2 ≥ b3 , and max(a1 + a2 − b1 , a3 )≥
max(b2 , b3 ) follows. We note that (a1 + a2 − b1 ) + a3 = b2 + b3 . Using the lemma yet another time
yields y a1 +a2 −b1 z a3 + y a3 z a1 +a2 −b1 ≥ y b2 z b3 + y b3 z b2 . Analogous to previous arguments one then
33
realizes that
X
X
xb1 (y b2 z b3 + y b3 z b2 ) also holds. Then we
(2)
xb1 (y a1 +a2 −b1 z a3 + y a3 z a1 +a2 −b1 ) ≥
cyclic
cyclic
note that
X
b1
b2 b3
b3 b2
x (y z + y z ) =
X
b1 b2 b3
x y z . At last we note that the right-hand side of (1)
sym
cyclic
is equal to the left-hand side of (2). Then, from (1) and (2), Muirhead’s inequality follows.
Case 2: b1 ≤ a2 .
According to the conditions, we have 3b1 ≥ b1 + b2 + b3 = a1 + a2 + a3 ≥ b1 + a2 + a3 . From
this follows that b1 ≥ a2 + a3 − b1 . According to the conditions, we also have a2 ≥ b1 . Hence,
max(a2 , a3 )≥ max(b1 , a2 + a3 − b1 ). We note that a2 + a3 = b1 + (a2 + a3 − b1 ). Now, we use
the lemma to get y a2 z a3 + y a3 z a2 ≥ y b1 z a2 +a3 −b1 + y a2 +a3 −b1 z b1 . Analogous with case 1, we see
without problem that
X
X
xa1 (y b1 z a2 +a3 −b1 + y a2 +a3 −b1 z b1 ) holds. We note that
(3)
xa1 (y a2 z a3 + y a3 z a2 ) ≥
X
a1
cyclic
a2 a3
x (y z
a3 a2
+y z )=
X
cyclic
a1 a2 a3
x y x .
sym
cyclic
According to the conditions, we also have max(a1 , a2 + a3 − b1 )≥ max(b2 , b3 ). We note that
a1 + (a2 + a3 − b1 ) = b2 + b3 and use the lemma one final time to get
xa1 z a2 +a3 −b1 + xa2 +a3 −b1 z a1 ≥ xb2 z b3 + xb3 z b2 . Finally we note that
X
X
(4)
y b1 (xa1 z a2 +a3 −b1 + xa2 +a3 −b1 z a1 ) ≥
y b1 (xb2 z b3 + xb3 z b2 ) holds.
cyclic
We see that
X
cyclic
cyclic
y b1 (xb2 z b3 + xb3 z b2 ) =
X
xb1 y b2 z b3 . Analogous with case 1 we see that the right-
sym
hand side of (3) is equal to the left-hand side of (4). Thus, from (3) and (4), Muirhead’s inequality
follows.
SOLUTIONS TO THE PROBLEMS 1
1. Using the hint, we realize that (x + y 3 )(y + z 3 )(z + x3 ) ≥ (x + 4y)(y + 4z)(z + 4x) =
(x + y + y + y + y)(y + z + z + z + z)(z + x + x + x + x). Then, applying the AM-GM inequality
to each of the parenthesis, we get (x + y + y + y + y)(y + z + z + z + z)(z + x + x + x + x) ≥
p
p
√
5
5 5 xy 4 · 5 5 yz 4 · 5 zx4 = 125xyz.
2. Applying the AM-GM inequality to the denominator of each term on the left-hand side of
x
y
x
y
1
1
1
the inequality, we get 4
+ 4
≤ p
+ p
=
+
=
.
2
2
x +y
y +x
2xy 2xy
xy
2 x4 y 2 2 y 4 x2
34
1 4 4
3. We apply the AM-GM inequality to the left-hand side of the inequality and get + + +
a b c
r
16
16
4 256
≥4
= √
. Then, using the AM-GM inequality again, this time on the denominator,
4
d
abcd
abcd
64
16
16
we get √
≥ a+b+c+d =
.
4
a+b+c+d
abcd
4
a b c
c b b a c a
4. We have 1 +
1+
1+
=1+ + + + + + +1=
b
c
a
a c a b b c
a a a b b b c c c
−1+ + + + + + + + + . Then, applying the AM-GM inequality to each of
a b c
b a c
c a b
a a a b b b c c c
a+b+c
the parenthesis, we get −1+ + + + + + + + +
≥ −1+3 √
. Now,
3
a b c
b a c
c a b
abc
a + b + c
a+b+c
all we have to show is that −1+3 √
≥
2
1+ √
, which is the same as showing that
3
3
abc
abc
a+b+c
√
≥ 3. This inequality is obviously true: it is just a restatement of the AM-GM inequality
3
abc
for three terms.
5. We start, as suggested, by showing that x2 + y 2 + z 2 ≥ xy + yz + zx. We√make √
the variable
√
substitution x2 = a, y 2 = b and z 2 = c. Then we have to show that a + b + c ≥ ab + bc + ca.
Using the AM-GM inequality on each of the terms on the right-hand side of the inequality, we get
√
√
√
a+b b+c c+a
ab + bc + ca ≤
+
+
= a + b + c. (Those familiar with the Rearrangement
2
2
2
inequality may think that the proof above is superfluous, since the proven inequality is a simple
consequence of the Rearrangement inequality.)
√
√
√
Now we can show that x2 + y 2 + z 2 + xy + yz + xz ≥ 2( x + y + z). By using our hint
inequality, we see that x2 + y 2 + z 2 + xy + yz + xz ≥ 2(xy + yz + xz) =
xy + xz xy + yz xz + yz 2
+
+
. Then we apply the AM-GM inequality to each term inside the
2
2
2
√
√
√
xy + xz xy + yz xz + yz parenthesis to get 2
+
+
≥ 2( x · xyz + y · xyz + z · xyz) =
2
2
2
p
√
√
√ √
√
√
√
√
√
2( 2x + 2y + 2z) = 2 2( x + y + z) ≥ 2( x + y + z).
6. We start with the hint 1 − yi =
X
yk . Using the AM-GM inequality on the right-hand
k6=i
sY
side of this inequality, we get 1 − yi ≥ (n − 1) n−1
yk . Of course this is true for all i (i =
k6=i
1, 2, ..., n), which means that we have n inequalities like the one above. We multiply together all
the right-hand sides of these n equalities and treat similarly all the left-hand sides. Then we come
n
n
n
Y
Y
Y
n
up with the inequality
(1 − yi ) ≥ (n − 1)
yi . Dividing both sides by
yi yields the
i=1
i=1
35
i=1
n
Y
1 − yi
1 − xi1998
1 − yi
xi + 1998 − 1998
xi
+1998
inequality
≥ (n − 1) . Now,
=
=
=
.
1998
yi
yi
1998
1998
xi +1998
i=1
n
Y
xi
≥ (n − 1)n , which after some easy rearranging becomes
Thus, our inequality becomes
1998
i=1
the inequality that we set out to prove.
n
7. For proof of the hint inequality, see problem 5. Using the AM-HM inequality on the left1
1
1
9
hand side of the inequality, we get
+
+
≥
. Then, applying
1 + ab 1 + bc 1 + ca
3 + ab + bc + ca
9
3
9
9
≥
the hint inequality, we get
= = .
2
2
2
3 + ab + bc + ca
3+a +b +c
6
2
1
≥ 2 (t > 0). We apply the AM-GM inequality to
t
r
1
1
the left-hand side of this inequality to get t + ≥ 2 t · = 2.
t
t
8. We start by showing the inequality t +
Now to the main inequality. We note that all the denominators are positive. Then we subtract
xn from both sides of the inequality and express the left-hand side of the inequality in a clever way.
We now have to prove the inequality
1
1
1
(x0 − x1 ) +
+ (x1 − x2 ) +
+ ... + (xn−1 − xn ) +
≥ 2n.
x0 − x1
x1 − x2
xn−1 − xn
This becomes easy with the help of our hint inequality. The left-hand side of the above inequality
consists of 2n terms. If we add the two first terms together, and then add the two following terms
together, and so on, then, according to our hint inequality, the above inequality holds.
9. We start by multiplying both sides of the equality by (4x2 + y 2 ). Then, using the AM-GM
inequality to each parenthesis on the right-hand side, we get
p
p
p
√
64x2 y 2 = (4x2 + y 2 )(x + 1)(y + 2)(2x + y) ≥ 2 4x2 · y 2 · 2 x · 2 2y · 2 2xy = 64x2 y 2 .
Hence the inequality above is an equality and the equality condition in AM-GM implies that
x = 1 and y = 2.
10. We follow the hint and apply the AM-GM inequality three times; to the left-hand side of the
equality on the factors (1+2ax) = (1+ax+ax) and (1+2by) = (1+by+by), and then r
to the whole
27xy
27xy
xy
p
right-hand side. We get the following:
≤ √
= 3 3 2 2 and
3
3
2 2
2 2
(1 + 2ax)(1 + 2by)
ab
r
r3 a x ·3 b y
1
x y
xy
27xy
xy
1
x y
+ + ≥ 3 3 2 2 . Hence
≤33 2 2 ≤
+ + . Now, to have
ab a
b
ab
(1 + 2ax)(1 + 2by)
ab
ab a
b
equality, the condition for equality in the AM-GM inequality has to be met. This condition leads
to
(1) ax = by = 1 and
(2)
1
x
y
= = .
ab
a
b
36
1
1
1
1
, y = while from (2), we get x = , y = . Hence, if we have a 6= b, then
a
b
b
a
1
x y
27xy
=
+ + . But if
there is no pair of positive real numbers x, y such that
(1 + 2ax)(1 + 2by)
ab a b
1
we have a = b, then we have equality for x = y = .
a
(1) yields x =
11. We start by multiplying both sides of the inequality by (an +bn +cn ). Because of symmetry,
we can, without loss of generality, assume that a ≤ b ≤ c. Then an ≥ bn ≥ cn and the Chebyshev’s
1
1
1
inequality implies (an + bn + cn ) · (a + b + c) ≤ (an+1 + bn+1 + cn+1 ), which is the inequality
3
3
3
in question.
12. We take logarithms on both sides and then use the well-known logarithm laws to rewrite
the inequality. Since logarithm is a strictly increasing function, the inequality is not affected. We
a+b+c
get a ln a + b ln b + c ln c ≥
(ln a + ln b + ln c), which is the the inequality we want to
3
prove.
Because of symmetry, we can once again without loss of generality assume that a ≤ b ≤ c.
Then the inequality is an immediate consequence of the Chebyshev’s inequality.
1
1
1
13. The left-hand side of the inequality can be rewritten as a1 · + a2 · 2 + ... + an · 2 . It
1
2
n
1
1
1
1
is obvious that ≥ 2 ≥ ... ≥ 2 . Then, according to the Rearrangement inequality, a1 · + a2 ·
1
2
n
1
1
1
+ ... + an · 2 is minimal when we have a1 < a2 < ... < an . Since all numbers ai are distinct,
22
n
1
1
1
1
1
1
1 1
1
then ai ≥ i. Thus we have a1 · +a2 · 2 +...+an · 2 ≥ 1· +2· 2 +...+n· 2 = + +· · ·+ .
1
2
n
1
2
n
1 2
n
1
1
1
+ x2 ·
+ ... + xn · . To
x2
x3
x1
minimize this expression with help from the Rearrangement inequality, the largest of the numbers
1 1
1
x1 , x2 , ..., xn has to be multiplied by the smallest of the numbers , , ..., , the second largest
x1 x2
xn
multiplied by the second smallest, and so on. Now, suppose that of all the numbers x1 , x2 , ..., xn ,
1 1
1
1
the number xk is the largest. Which one of the numbers , , ...,
is then the smallest?
of
x1 x2
xn
xk
1
course! If the number xl is the second largest then
is the second smallest, and so on. Thus
xl
1
1
1
1
1
1
+ a2 ·
+ ... + an ·
≥ a1 ·
+ a2 ·
+ ... + an ·
= n.
a1 ·
a2
a3
a1
a1
a2
an
14. The left-hand side of the inequality can be rewritten as x1 ·
1
1
1
a8 + b8 + c8
5
5
5
=
a
·
+
b
·
+
c
·
. Because of the symmetry, we
a3 b3 c3
b3 c 3
a3 c 3
a3 b 3
can without loss of generality assume that a ≥ b ≥ c. From that follows that a5 ≥ b5 ≥ c5 and
1
1
1
≤ 3 3 ≤ 3 3 . Then, using the Rearrangement inequality twice, we get
3
3
ab
ac
bc
15. We have
37
a5 ·
a2 ·
1
b3 c 3
+ b5 ·
1
a3 c 3
+ c5 ·
1
a3 b 3
≥ a5 ·
1
a3 b 3
+ b5 ·
1
b3 c 3
+ c5 ·
1
a3 c 3
= a2 ·
1
1
1
+ b2 · 3 + c 2 · 3 ≥
3
b
c
a
1
1
1 1 1
1
2
2
+ + .
+
b
·
+
c
·
=
a3
b3
c3
a b c
16. By the Rearrangement inequality we have bz + cy ≤ by + cz. Hence (by + cz)(bz + cy) ≤
(by + cz) = b2 y 2 + 2bcyz + c2 z 2 . Moreover, since 2bcyz ≤ b2 y 2 + c2 z 2 then (by + cz)(bz + cy) ≤
a2 x 2
a2 x 2
2(b2 y 2 + c2 z 2 ). From this follows that
≥
.
(by + cz)(bz + cy)
2(b2 y 2 + c2 z 2 )
A similar argument for the two other terms on the left hand side, together with the substitution
1
α
β
γ +
+
.
α = a2 x2 , β = b2 y 2 , γ = c2 z 2 proves that the left-hand side is ≥
2 β+γ γ+α α+β
β
γ
3
α
+
+
≥ holds.
What we would need to show now is that the inequality
β+γ γ+α α+β
2
To that end multiply both sides of this inequality by (the positive number) α(β + γ) + β(γ + α) +
γ(α + β). Then, using the Cauchy-Schwarz inequality on the left-hand side , we get
α
β
γ +
+
· α(β + γ) + β(γ + α) + γ(α + β) ≥
β+γ γ+α α+β
s
r
r α
2
p
p
p
β
γ
· α(β + γ) +
· β(γ + α) +
· γ(α + β) =
β+γ
γ+α
α+β
(α + β + γ)2 = α2 + β 2 + γ 2 + 2(αβ + αγ + βγ).
On the right hand side of the inequality we have
3
α(β + γ) + β(γ + α) + γ(α + β) =
2
3(αβ + βγ + αγ).
This means that what remains to show is that α2 +β 2 +γ 2 +2(αβ+αγ+βγ) ≥ 3(αβ+βγ+αγ),
i.e. that α2 + β 2 + γ 2 ≥ αβ + βγ + αγ. For a simple proof, see problem 5.
17. Using the hint we note that (a + b) + (b + c) + (c + d) + (d + a) = 2 and we multiply both
sides of the inequality by this factor. We get the inequality
a2
b2
c2
d2 (a + b) + (b + c) + (c + d) + (d + a)
+
+
+
≥ 1.
a+b b+c c+d d+a
Using the Cauchy-Schwarz inequality on the left-hand side yields (a + b) + (b + c) + (c +
r
r
√
a2
b2
c2
d2 √
a2
b2
d) + (d + a)
+
+
+
≥
a+b·
+ b+c·
+
a+b b+c c+d d+a
a+b
b+c
r
r
√
√
c2
d2 2
= (a + b + c + d)2 = 1.
c+d·
+ d+a·
c+d
d+a
1
To realize that we have equality if and only if a = b = c = d = , we start by noting that when
4
we have equality, the number sequences r
r
r
r
√
√
√
√
a2
b2
c2
d2 a + b, b + c, c + d, d + a and
,
,
,
must be propora+b
b+c
c+d
d+a
tional. Thus when we have equality there is (and since all the numbers in these sequences are posi-
38
√
r
√
a2
, k b+c=
a+b
r
√
b2
, k c+d=
b+c
r
c2
tive) a number k > 0 such that k a + b =
c+d
r
√
d2
and k d + a =
.
d+a
These four equations yield in turn k = a = b = c = d. Hence, when we have equality, we have
1
a=b=c=d= .
4
18. We start, as suggested,
by multiplying both sides of the inequality by the factor a(b +
2c) + b(c + 2a) + c(a + 2b) . We have then to prove that
a
b
c a(b + 2c) + b(c + 2a) + c(a + 2b) ·
+
+
≥
b + 2c c + 2a a + 2b
a(b + 2c) + b(c + 2a) + c(a + 2b).
Let us apply the Cauchy-Schwarz inequality the the left-hand side of this inequality. We get
a
b
c a(b + 2c) + b(c + 2a) + c(a + 2b) ·
+
+
≥
b + 2c c + 2a a + 2b
r
r
r
p
p
p
a
b
c 2
+ (b(c + 2a) ·
+ c(a + 2b) ·
a(b + 2c) ·
=
b + 2c
c + 2a
a + 2b
(a + b + c)2 = a2 + b2 + c2 + 2(ab + ac + bc).
Working out the right-hand side of the inequality, we get a(b + 2c) + b(c + 2a) + c(a + 2b) =
3(ab + ac + bc). Once again we end up having to prove the inequality
a2 + b2 + c2 ≥ ab + ac + bc. See problem 5 for that proof.
19. We may begin by multiplying both sides of the inequality by (ac + bd) to get the inequality
a3 b 3 +
≥ ac + bd. Then, applying the Cauchy-Schwarz inequality to the left-hand
(ac + bd)
c
d
r
r a3 b 3 √
a3 √
b3 2
side yields (ac + bd)
+
≥
ac ·
+ bd ·
= (a2 + b2 )2 .
c
d
c
d
We now have to show that the inequality (a2 + b2 )2 ≥ ac + bd holds. We square this inequality.
Then we get (a2 + b2 )4 = (a2 + b2 )(a2 + b2 )3 = (a2 + b2 )(c2 + d2 ) ≥ (ac + bd)2 . The inequality
(a2 + b2 )(c2 + d2 ) ≥ (ac + bd)2 holds, of course, according to the Cauchy-Schwarz inequality.
20. For this problem we will give two solutions.
Solution 1 (using Hölder’s inequality): We make the suggested variable substitution. The inequality to be proven becomes (1 + an1 )(1 + an2 )...(1 + ann ) ≥ (1 + a1 a2 ...an )n .
1
If we now take the n:th root of both sides and then write the inequality as (1n + an1 ) n (1n +
1
1
an2 ) n ...(1n + ann ) n ≥ (1 + a1 a2 ...an ). It is obvious that this inequality holds (Hölder’s inequality
for n number sequences).
Solution 2 (using the AM-GM inequality): Using the binomial theorem, we develop both sides
of the inequality. We get the inequality
39
1 + (x1 + x2 + ... + xn ) + (x1 x2 + x1 x3 + ... + xn−1 xn ) + ... + (x1 x2 ...xn ) ≥
1
2
n
n
n
1+
(x1 x2 ...xn ) n +
(x1 x2 ...xn ) n + ... +
(x1 x2 ...xn ).
1
2
n
Let us subtract 1 from both sides of the inequality. Then we have n parenthesis on the left-hand
side of the inequality, and n terms on the right-hand side. If we are able to show that the k:th
parenthesis on the left-hand side (k = 1, 2, ..., n) is ≥ the k:th term on the right-hand side, we
n
are done. Now, the k:th parenthesis on the left-hand side consists of
terms. The factor xi ,
k
n−1
n
(k = 1, 2, ..., n) exists in
of these terms. We use the AM-GM inequality on these
k−1
k
n−1 n−1
n−1 1/(n)
n
k
( ) ( ) ( )
.
terms to get that the parenthesis is ≥
x1 k−1 x2 k−1 ...xnk−1
k
n−1
(n − 1)!
n
k!(n − k)!
We note that
=
, and that 1/
=
=
k−1
(k − 1)!(n − k)!
k
n!
n−1 n−1
n
k (k − 1)!(n − k)!
n
( ) ( ) (n−1) 1/(k )
=
·
. From this it follows that
x1 k−1 x2 k−1 ...xnk−1
n
(n − 1)!
k
k
k
n
n
(x1 x2 ...xn ) n . Since
(x1 x2 ...xn ) n is the k:th term on the right-hand side of the inequak
k
lity, we are done.
2
1
21. Using Hölder’s inequality (with p = and q = ) on the right-hand side of the inequality,
3
2 3
1
2 23
2 32 3 3
2
2
2
2 23
(b + c3 + a3 ) 3 = a3 + b3 + c3 .
we get a b + b c + c a ≤ (a ) + (b ) + (c )
22. Applying Hölder’s inequality for three number sequences (these sequences being (ap , bp , cp ),
and (aq , bq , cq ), (ar , br , cr )) to the right-hand side of the inequality yields ap bq cr +cp aq br +bp cq ar ≤
1
1
1
1 p
1
1 q
1
1
1 r
(ap ) p + (bp ) p + (cp ) p · (bq ) q + (aq ) q + (cq ) q · (cr ) r + (br ) r + (ar ) r =
(a + b + c)(p+q+r) = a + b + c.
23. We start by taking the third root of both sides of the inequality. Then we express xk and
n
1 23 32
X
13
1
1
1
1
3 3
as (xk ) and
respectively. We come up with the inequality 2 3
(xk3 )3
·
xk
xk
k=1
n X
1 23 32 32
≥ n. Using first Hölder’s inequality on the left-hand side of this inequality and
x
k
k=1
n
n X
31 X
1
1
1 23 32 32
3
3
(xk )
then using the condition xk ∈ [1, 2], k = 1, 2, ..., n, we get 2 3
·
≥
x
k
k=1
k=1
n n
n
X
X
X
1 32 1
1
1
1
1
1
≥ 23
xk3 ·
= 23
≥ 23
= n.
1
1
x
3
3
k
2
x
k=1
k=1
k=1
k
40
√
√
√
y b
z c
x a
, a2 = √
, a3 = √
,
24. We set a1 = √
a+b+c
a+b+c
a+b+c
√
√
√
y a
z b
x c
b1 = √
, b2 = √
, b3 = √
and
a+b+c
a+b+c
a+b+c
√
√
√
z a
x b
y c
c1 = √
, c2 = √
, c3 = √
,
a+b+c
a+b+c
a+b+c
and use Minkowski’s inequality for three number sequences (a1 , a2 , a3 ), (b1 , b2 , b3 ), (c,1 , ,2 , c3 )
and r = 2. We get
r
r
r
q
ax2 + by 2 + cz 2
ay 2 + bz 2 + cx2
az 2 + bx2 + cy 2
+
+
= a21 + a22 + a23 +
a+b+c
a+b+c
a+b+c
q
q
p
b21 + b22 + b23 + c21 + c22 + c23 ≥ (a1 + b1 + c1 )2 + (a2 + b2 + c2 )2 + (a3 + b3 + c3 )2 =
s √
x a + y √ a + z √ a 2 y √ b + z √ b + x √ b 2 z √c + x √ c + y √c 2
√
√
√
+
+
,
a+b+c
a+b+c
a+b+c
which after some simplification becomes x + y + z.
25. We rearrange the inequality and get the equivalent expression to prove
r 1 1 4 16 √
2
+ + +
· a + b + c + d ≥ 64.
a b c
d
Now we can use Hölder’s inequality on the left-hand side to get
r
1 1 4 16 √
2 1 √
1 √
2 √
4 √ 2
+ + + · a + b + c + d ≥ √ · a+ √ · b+ √ · c+ √ · d = 82 = 64.
a b c
d
a
c
b
d
1
26. First we show that the left inequality holds. We note that f (x) = is strictly convex for
x
2
00
x > 0 (we have f (x) = 3 > 0 for x > 0). Then we divide both sides of the inequality by 6
x
1, 5
1
1
1
1
1
1
to come up with the inequality
≤ ·
+ ·
+ ·
. Using Jensen’s
a+b+c
3 a+b
3 b+c
3 c+a
inequality (with the function mentioned above) on the right-hand side of the inequality then yields
1
1
1
1
1
1
1
3
1, 5
·
+ ·
+ ·
≥ 1
=
=
.
1
1
3 a+b 3 b+c 3 c+a
a+b+c
(a + b) + 3 (b + c) + 3 (c + a) 2a + 2b + 2c
3
To show that the right-hand inequality holds we start by expressing the right-hand side of this
inequality in a suitable way. Then, once again, we use Jensen’s inequality; this time on each of the
1 1 1 1 1 1 1
parenthesis (the function being the same as before). We get + + =
· + ·
+
a b c
2 a 2 b
1 1 1 1 1 1 1 1
1
1
1
1
1
1 · + ·
+
· + ·
≥ 1
+
+
=
2
+
+
.
1
1
1
1
1
2 b 2 c
2 c 2 a
a
+
b
b
+
c
c
+
a
a
+
b
b
+
c
c
+
a
2
2
2
2
2
2
27. We begin by taking a look at the left-hand side of the inequality. We rearrange this expres-
41
x
y
z
x+1−1 y+1−1 z+1−1
1
1
+
+
=
+
+
= 3−
−
−
x+1 y+1 z+1
x+1
y+1
z+1
x+1 y+1
1
1
1
3
1
. Our inequality to be proven is thus the following: 3 −
−
−
≤ . After
z+1
x+1 y+1 z+1
4
some easy rearranging we come up with the equivalent inequality
3
1 1
1
1 1
≤
+
+
. We note once more that f (x) =
is strictly convex for
4
3 x+1
y+1
z+1
x
x > 0. Then we apply Jensen’s inequality to the right-hand side of the last inequality and get
1 1
1
1 1
3
3
+
+
≥ 1
=
= .
1
1
3 x+1 y+1 z+1
3+x+y+z
4
(x + 1) + 3 (y + 1) + 3 (z + 1)
3
sion, getting
28. We set f (x) =
1
. If one takes a look on f 00 (x), one realizes that we have f 00 (x) =
4x − x3
12(x2 − 1)2 + 20
> 0 for 0 < x < 2, which means that f (x) is strictly convex in this interval. We
x3 (4 − x2 )3
n
n
X
X
1
1
1
=
n
·
·
≥
can then use Jensen’s inequality. We get the following:
3
3
4a
n
4a
k − ak
k − ak
k=1
k=1
n
n
n
P
P
= 1
= . Thus, the minimum value of the
3
4( n1 · nk=1 ak ) − ( n1 · nk=1 ak )3
4( n · n) − ( n1 · n)3
n
expression is and is attained when a1 = a2 = ... = an = 1 .
3
29. If one draws an arbitrary triangle with an inscribed circle of radius 1 it is not difficult to
realize that the hint holds.
1
π
Now, consider the function f (x) =
. It is strictly convex for 0 < x < , which one
tan x
2
π
sin 2x
00
> 0 for 0 < x < .
realizes by studying f (x) =
2
sin4 x
1
1
1
1
1
1
1
Using Jensen’s inequality then yields
+
+
=3
·
+ ·
+
tan
α
tan
β
tan
γ
3
tan
α
3
tan
β
1 3
1
3
1
1
1
·
≥
=
,
π =
π +
π +
1
3 tan γ
tan 6
tan 6
tan 6
tan π6
tan( 3 (α + β + γ))
which is half the circumference of the equilateral triangle.
30. We make the suggested variable substitution to get the inequality
r
r
ea
eb
ec
+
+
≥ 1. The conditions 0 < x, y, z < 4 and xyz = 1 become the
ea + 8
eb + 8
ec + 8
conditions 0 < ea , eb , ec < 4 and a + b + c = 0 respectively. Then we study the function
s
s
es 21
4e 2
2e 2 (8 − 2es )
0
00
f (s) =
. After some work we get f (s) =
. We
3 and f (s) =
5
es + 8
(es + 8) 2
(es + 8) 2
see that f 00 (s) > 0 for es < 4. Hence f (s) is strictly convex
for es < 4. Now we multiply both
r
r
r
1
1
ea
1
eb
1
ec
1
sides of our inequality by , and we get
+
+
≥
.
3
3 ea + 8 3 eb + 8 3 ec + 8
3
Applying now Jensen’s inequality to the left-hand side yields
r
42
1
3
r
ea
1
+
a
e +8 3
r
eb
1
+
b
e +8 3
r
ec
ec
+8
s
≥
1
e 3 (a+b+c)
e
1
(a+b+c)
3
+8
r
=
1
1
= , and we are done.
1+8
3
31. We use the hint and find, according to the Power Mean inequality, that
x2 + y 2 12 x3 + y 3 13
2
1
≤
, which is equivalent to the inequality x2 + y 2 ≤ (x3 + y 3 ) 3 · 2 3 .
2
2
Applying now the given condition x3 + y 3 > 2 to the right-hand side of this inequality yields
2
1
2
1
(x3 + y 3 ) 3 · 2 3 < (x3 + y 3 ) 3 · (x3 + y 3 ) 3 = x3 + y 3 . Thus we have
x2 + y 2 < x 3 + y 3 .
(1)
We also have (1−y)2 ≥ 0. Developing the left-hand side of this inequality and then multiplying
both sides by y 2 yield
y 2 − 2y 3 + y 4 ≥ 0.
(2)
Now, if we add the left-hand and the right-hand sides of (1) and (2) respectively, we come up
with the desired inequality.
32. We start by showing that the inequality
4
X
i=1
a3i ≥
4
X
ai . holds. To this end we divide
i=1
both sides of the inequality by 4. Then we apply the Power Mean inequality to the right-hand side,
13
1
1
3
3
3
3
getting (a1 + a2 + a3 + a4 ) ≤
(a1 + a2 + a3 + a4 ) . What we now have to show is that
4
4
31
1 3
1 3
(a1 + a2 3 + a3 3 + a4 3 ) ≥
(a1 + a2 3 + a3 3 + a4 3 ) . It is obvious that this inequality holds
4
4
if the inequality
1 3
a1 + a2 3 + a3 3 + a4 3 ≥ 1 holds. And it does: using the AM-GM inequality on the left-hand
4
p
1
side yields (a1 3 + a2 3 + a3 3 + a4 3 ) ≥ 4 a1 3 · a2 3 · a3 3 · a4 3 = 1.
4
4
4
X
X
1
3
The second inequality that we have to show is
ai ≥
. To show that, we set A =
a
i=1
i=1 i
4
4
X
X
a1 3 +a2 3 +a3 3 +a4 3 and Ai = A−ai 3 , (i = 1, 2, 3, 4). We see then that
Ai = 4A−
a3i = 3A.
i=1
Thus,
4
1X
3
Ai = A. Now we note that, by the AM-GM inequality, we have
i=1
i=1
1
1
A1 = (a2 3 +
3
3
1
. Using the very same reasoning, we realize that the
a1
1
1 1
1
1
1
inequalities A2 ≥ , A3 ≥
and A4 ≥
also hold. Adding the left-hand and the right3
a2 3
a3
3
a4
hand sides respectively of these four inequalities then yields the desired inequality.
3
3
a3 + a4 ) ≥
p
3
a2 3 a3 3 a4 3 = a2 a3 a4 =
33. We begin by showing that a3 + b3 + c3 ≥
43
1
(a + b + c)(a2 + b2 + c2 ). We multiply both
3
1
sides of this inequality by and apply the Power Mean inequality to the right-hand side of the
3
1
1
inequality then yields (a + b + c) · (a2 + b2 + c2 ) ≤
3
1
313 1
23
1
3
3
3
2 32
2 32
2 23
≤
(a + b + c ) ·
((a ) + (b ) + (c ) ) = (a3 + b3 + c3 ).
3
3
3
Now let’s turn to the inequality that we are supposed to show. Using the recently proved inequality on the left-hand side the inequality to be shown, we get
1
(a + b + c)(a2 + b2 + c2 )
a3 + b3 + c3 a3 + b3 + d3 a3 + c3 + d3 b3 + c3 + d3
3
+
+
+
≥
+
a+b+c
a+b+d
a+c+d
b+c+d
a+b+c
1
1
1
(a + b + d)(a2 + b2 + d2 )
(a + c + d)(a2 + c2 + d2 )
(b + c + d)(b2 + c2 + d2 )
3
+ 3
+ 3
=
a+b+d
a+c+d
b+c+d
1 2
1
1
1
(a + b2 + c2 ) + (a2 + b2 + d2 ) + (a2 + c2 + d2 ) + (b2 + c2 + d2 ) = a2 + b2 + c2 + d2 .
3
3
3
3
34. We follow the hint
the inequality. On the left-hand side of the inequality
p and homogenize
√
we multiply the term 2 3xyz by 1 = x + y + z. Then we multiply the right-hand side by
1p= (x + y + z)2 . We develop the right-hand side and simplify to come up with the inequality
√
3xyz · x + y + z ≤ xy + xz + yz. We continue by squaring both sides and then getting rid
of terms that cancel each other. We get the inequality x2 yz + xy 2 z + xyz 2 ≤ x2 y 2 + x2 z 2 + y 2 z 2 .
To see that this inequality holds, we make the variable substitution xy = a, xz = b, yz = c. This
leads to the, probably by now, familiar inequality ab + ac + bc ≤ a2 + b2 + c2 , which holds. For
proof of that, see problem 5.
35. In order to use Schur’s inequality we need to make the given inequality homogeneous. This
is done by multiplying the left-hand side of the inequality by 1 = (p + q + r) and the term 2 on
the right-hand side by 1 = (p + q + r)3 . Multiplying out and canceling some terms we end up with
p2 q + pq 2 + p2 r + pr2 + q 2 r + qr2 ≤ 2(p3 + q 3 + r3 ).
Rewriting the right-hand side of this expression and then using the AM-GM inequality
on the
p
3
3
3
3
3
3
3
3
3
3
3
3
3
3
parenthesis yields 2(p + q + r ) = p + q + r + (p + q + r ) ≥ p + q + r + 3 p q 3 r3 =
p3 + q 3 + r3 + 3pqr. What is now left for us to show is that the inequality p2 q + pq 2 + p2 r + pr2 +
q 2 r + qr2 ≤ p3 + q 3 + r3 + 3pqr holds. Not much to show though, since this is Schur’s inequality
for r = 1.
n
36. We use the suggested notation and get (1 + a1 )(1 + a2 ) · ... · (1 + an ) = 1 +
S1 +
1
n
n
S2 + ... +
Sn .
2
n
1
n
n
n
n n1
n n
Now we will show that 1 +
S1 +
S2 + ... +
Sn ≥ (1 + Sn ) = .1 +
Sn +
1 2
n
1
n−1
n n2
n
n
Sn + ... +
Sn n +
Sn . In fact we find that for all k = 0, 1, 2, ..., n we have
n
2
n−1
1
1
n
n nk
Sk ≥
Sn . This is true, since, by MacLaurin’s inequality, Skk ≥ Snn .
k
k
44
1
Now, what we have shown is that 2n = (1 + a1 )(1 + a2 ) · ... · (1 + an ) ≥ (1 + (a1 · a2 · ... · an ) n )n .
Taking the n:th root of both sides, subtracting 1 and last taking the n:th power yields the desired
inequality.
A maybe easier solution uses Hölder’s inequality, as in problem 20.
37. We make the suggested substitutions a = x3 , b = y 3 , c = z 3 , multiply both sides by x3 y 3 z 3
to
Xget rid of the
Xdenominators, multiply out the parenthesis and we end up with the inequality
6 3 0
xy z ≥
x5 y 2 z 2 . This is clearly true by Muirhead’s inequality for a1 = 6, a2 = 3, a3 = 0,
sym
sym
b1 = 5, b2 = 2, b3 = 2.
38. In order to get rid of the denominators we start by multiplying both sides by 4(x + y)(x +
z)(y + z)(x + y + z). Then we multiply the parenthesis and cancel terms that occur on both sides.
Eventually we
get the inequality
6xyz ≤ x2 y + x2 z + y 2 x + y 2 z + z 2 x + z 2 y, which also can be
X
X
expressed as
xyz ≤
x2 yz 0 . It is easy to see that this inequality holds since it is Muirhead’s
sym
sym
inequality for a1 = 2, a2 = 1, a3 = 0, b1 = b2 = b3 = 1.
39. By making the trigonometric substitution x = cos α, y = cos β, for 0 ≤ α, β ≤ π, and
simplifying the expression using well-known trigonometric identities, we get
p
√
f (cos α, cos β) = cos α cos β + cos α 1 − cos2 β + cos β 1 − cos2 α−
q
q
p
p
2
2
2
2
(1 − cos α)(1 − cos β) = cos α cos β + cos α sin β + cos β sin α − sin2 α sin2 β =
cos α cos β + cos α sin β + cos β sin α − sin α sin β = cos(α + β) + sin(α + β) =
√ π
√ 1
1
π
2 √ cos(α + β) + √ sin(α + β) = 2 sin cos(α + β) + cos sin(α + β) =
4
4
2
2
√
π
√
+ α + β ≤ 2.
2 sin
4
π
The equality is attained for all α, β. such that α+β = . Thus, for real numbers −1 ≤ x, y ≤ 1,
4
√
the function f (x, y) has the maximum value 2 and the maximum is attained for all x = arccos α,
π
y = arccos β such that α + β = .
4
α
β
γ
40. We make the suggested variable substitution x = tan , y = tan and z = tan ,
2
2
2
where 0 < α, β, γ < π. Using some well-known trigonometric identities we eventually reduce the
inequality to
sin 2α + sin 2β + sin 2γ ≤ sin α + sin β + sin γ.
α
β
β
γ
γ
α
Now, the condition xy + yz + zx = 1 becomes tan tan + tan tan + tan tan = 1.
2
2
2
2
2
2
α
β
γ
Multiplying both sides of this equality by cos cos cos and then some simplifying give us the
2
2
2
α+β+γ
equality cos
= 0. Thus, the condition α + β + γ = π is in this case equivalent to the
2
(1)
45
condition xy + yz + zx = 1.
Bearing in mind that α + β + γ = π we express the left-hand side of (1) in another way, using
trigonometric identities. We get
sin 2α + sin 2β + sin 2γ = 2 sin(α + β) cos(α − β) + sin 2γ = 2 sin γ cos(α − β) + 2 sin γ cos γ =
2 sin γ(cos(α−β)+cos γ) = 2 sin γ(cos(α−β)−cos(α+β)) = 2 sin γ(cos α cos β +sin α sin β−
α
β
γ
α
β
γ
sin sin cos cos cos .
2
2
2
2
2
2
Then we treat the right-hand side of (1) in a similar way and we get
α−β
α+β
α−β
α+β
cos
+ sin(α + β) = 2 sin
cos
+
sin α + sin β + sin γ = 2 sin
2
2
2
2
α+β
α+β
α + β
α−β
α + β
2 sin
cos
= 2 sin
cos
+ cos
=
2
2
2
2
2
α + β
α
β
α
β
α
β
α
β
2 sin
cos cos + sin sin + cos cos − sin sin
=
2
2
2
2
2
2
2
2
2
π γ α+β
α
β
α
β
α
β
γ
4 sin
cos cos = 4 sin
−
cos cos = 4 cos cos cos .
2
2
2
2
2
2
2
2
2
2
Hence, our inequality to be proven becomes
β
γ
α
β
γ
α
β
γ
α
32 sin sin sin cos cos cos ≤ 4 cos cos cos , which, after dividing both sides of
2
2
2
2
2
2
2
2
2
α
β
γ
the inequality by 4 cos cos cos simplyfies to
2
2
2
β
γ
α
(2) 8 sin sin sin ≤ 1.
2
2
2
Now, since we have 0 < α, β, γ < π, we consider the function f (x) = sin x. We note that
00
f (x) = − sin x < 0 for 0 < x < π. Then we first apply the AM-GM inequality to the left-hand
side of (2) and after that Jensen’s inequality (remembering that Jensen’s inequality in this case is
reversed). We find that
1
1 α
β
γ
α
1
β
1
γ 3
1 β
1 γ
α
8 sin sin sin ≤ 8 sin + sin + sin
≤ 8 sin3
· + · + ·
=
2 2
2
3
2
3
2
3
2
3 2
3 2
3 2
π
8 sin3
= 1.
6
cos α cos β + sin α sin β) = 4 sin α sin β sin γ = 32 sin
PROBLEMS 2
41. Prove the inequality
an
bn
cn
an−1 + bn−1 + cn−1
+
+
≥
.
b+c c+a a+b
2
42. Let x1 , x2 , ..., xn be positive real numbers. Prove that
46
xx1 1 xx2 2 ...xxnn ≥ (x1 x2 ...xn )
x1 +x2 +...+xn
n
.
(x + y)(1 − xy)
1
1
≤ .
43. Prove that, for any real numbers x, y, − ≤
2
2
2
(1 + x )(1 + y )
2
44. (IMO, 1995) Let a, b and c be positive real numbers such that abc = 1. Prove that
1
1
1
3
+ 3
+ 3
≥ .
3
a (b + c) b (a + c) c (a + b)
2
45. (Moldavian MO, 1996) Let a, b and c be positive integers such that a2 + b2 − ab = c2 .
Prove that (a − c)(b − c) ≤ 0.
46. (Ireland, 2000) Let x and y be nonnegative real numbers such that x + y = 2. Prove that
x y (x2 + y 2 ) ≤ 2.
2 2
47. (Thailand, 1991) Let a, b, c and d be positive real numbers such that ab + bc + cd + da = 1.
a3
b3
c3
d3
1
Prove that
+
+
+
≥ .
b+c+d c+d+a d+a+b a+b+c
3
48. (Greece, 1987) Let a, b and c be positive real numbers. Show that for every integer n ≥ 1,
2 n−2 a + b + c n−1
an
bn
cn
+
+
≥
holds.
b+c c+a a+b
3
2
49. (Belarus, 1993) Let x1 , x2 , ..., xn (n ≥ 2) be nonnegative numbers such that x1 + x2 + ... +
n
X
p
√
xn = 1. Prove that
xi (1 − xi ) ≤ n − 1.
i=1
50. (Hungary, 1985) Let a, b, c and d be positive real numbers. Prove that
c
d
+
≥ 2.
d+a a+b
p
a
b
+
+
b+c
c+d
51. (Leningrad, 1981) Let a, b and c be real numbers such that 0 ≤ a, b, c ≤ 1. Prove that
p
p
√
a(1 − b)(1 − c) + b(1 − a)(1 − c) + c(1 − a)(1 − b) ≤ 1 + abc.
52. (Romania, 2004) Find all positive real numbers a, b, c which satisfy the inequality
4(ab + bc + ca) − 1 ≥ a2 + b2 + c2 ≥ 3(a3 + b3 + c3 ).
53. (Romania, 2004) Let a, b and c be real numbers such that a2 + b2 + c2 = 3. Prove that
|a| + |b| + |c| − abc ≤ 4.
54. (Estonia, 2004) Let a, b and c be positive real numbers such that
1
1
1
a + b2 + c2 = 3. Prove that
+
+
≥ 1.
1 + 2ab 1 + 2bc 1 + 2ca
2
47
55. (Austria, 2004) Let a, b, c and d be real numbers. Prove that
6
a + b6 + c6 + d6 − 6abcd ≥ −2.
56. (Ireland, 2004) Let a and b be nonnegative real numbers. Prove that
√ p
√
2( a(a + b)3 + b a2 + b2 ) ≤ 3(a2 + b2 ), with equality if and only if a = b.
57. (New Zealand, 2004) Let x1 , x2 , y1 , y2 be positive real numbers. Prove that
x21
y1
p
+
x22
(x1 + x2 )2
≥
.
y2
y1 + y2
58. Let a, b and c be real numbers. Prove that
√
p
p
3 2
2
2
2
2
2
2
a + (1 − b) + b + (1 − c) + c + (1 − a) ≥
.
2
59. Let a, b and c be positive real numbers such that abc = 1. Prove that
√
√
b+c c+a a+b √
√ + √ + √ ≥ a + b + c + 3.
a
c
b
60. Let a, b and c be positive real numbers. Prove that
b
c
9
a
+
+
≥
.
2
2
2
(b + c)
(c + a)
(a + b)
4(a + b + c)
61. (Juniors Balkan
√MO, 2002)
√ c be positive real numbers such that abc = 2. Prove
√ Let a, b and
that a3 + b3 + c3 ≥ a b + c + b c + a + c a + b.
62. Let a, b and c be positive real numbers such that abc ≤ 1. Prove that
a b c
+ + ≥ a + b + c.
b c a
63. (Juniors Balkan MO, 2002) Let a, b and c be positive real numbers. Prove that
a2 b 2 c 2
c3
≥
+ + .
a2
b
c
a
a3 b 3
+ 2+
b2
c
64. Let x, y and z be positive real numbers such that x2 + y 2 + z 2 + 2xyz = 1. Prove that
3
x+y+z ≤ .
2
65. (Juniors Balkan MO, 2003) Let x, y and z be real numbers such that
1 + x2
1 + y2
1 + z2
x, y, z > −1. Prove that
+
+
≥ 2.
1 + y + z 2 1 + z + x2 1 + x + y 2
66. Let a, b and c be positive real numbers such that a + b + c = 1. Prove that
2
a + b b2 + c c 2 + a
+
+
≥ 2.
b+c
c+a
a+b
48
67. (IMO, 1987) Let x, y and z be real numbers such that x2 + y 2 + z 2 = 2. Prove that
x + y + z ≤ xyz + 2.
68. Let x, y, z and α be positive real numbers such that xyz = 1 and α ≥ 1.
xα
yα
zα
3
Prove that
+
+
≥ .
y+z z+x x+y
2
69. Let a, b, c and d be positive real numbers. Prove that
(a + b)3 (b + c)3 (c + d)3 (d + a)3 ≥ 16a2 b2 c2 d2 (a + b + c + d)4 .
70. Let a, b and c be positive real numbers such that a + b + c = 1. Prove that
(a2 + b2 )(b2 + c2 )(c2 + a2 ) ≥ 8(a2 b2 + b2 c2 + c2 a2 )2 .
ONE MORE USEFUL INEQUALITY
The AM-GMinequality can easily begeneralized to matrices with nonnegative elements.
a11 a12 ... a1n
 a21 a22 ... a2n 
 be a matrix with nonnegative elements. Let G1 , G2 , ..., Gm
Let A = 
 ...
... ... ... 
am1 am2 ... amn
denote the geometric means of the rows of A and let A1 , A2 , ..., An be the arithmetic means of the
columns of A. Suppose that all the numbers A1 , A2 , ..., An are positive.
Consider one of the rows of A, let’s say the first row. After dividing a1k by Ak (k = 1, 2, ..., n),
a1n
a11 a12
we get a list of n numbers:
,
, ...,
.
A 1 A2
An
Applying now the AM-GM inequality to those numbers yields
r
n
1 X a1k
a11 a12 ...a1n
G1
≥ n
=
,
n k=1 Ak
A1 A2 ...An
G(A1 , A2 , ..., An )
where G(A1 , A2 , ..., An ) denotes the geometric mean of A1 , A2 , ..., An .
After repeating this procedure for all rows of A, we may add all m inequalities. This will give
m
n
m
1 X X ajk X
Gj
≥
.
n j=1 k=1 Ak
G(A1 , A2 , ..., An )
j=1
After rearranging the terms on the left-hand side we will get
Pm
n Pm
1 X j=1 ajk
j=1 Gj
≥
,
n k=1
Ak
G(A1 , A2 , ..., An )
49
Pm
Pm
n
1
1X
j=1 Gj
j=1 Gj
m
, i.e. 1 ≥
. If we let
m ≥
which is equivalent to
n k=1
G(A1 , A2 , ..., An )
G(A1 , A2 , ..., An )
A(G1 , G2 , ..., Gm ) be the arithmetic mean of G1 , G2 , ..., Gm , then the obtained inequality can
be written as G(A1 , A2 , ..., An ) ≥ A(G1 , G2 , ..., Gm ) . It should be obvious that this inequality
is valid even if we allow some of the Ak to be 0. Hence we have proved the following, powerful
generalization of the AM-GM inequality:
Theorem 4. If A is a matrix with m rows, n columns and nonnegative elements, let G1 , G2 , ..., Gm
denote the geometric means of the rows of A and let A1 , A2 , ..., An be the arithmetic means of the
columns of A. Moreover, let G(A1 , A2 , ..., An ) denote the geometric mean of A1 , A2 , ..., An and
let A(G1 , G2 , ..., Gm ) be the arithmetic mean of G1 , G2 , ..., Gm . Then
G(A1 , A2 , ..., An ) ≥ A(G1 , G2 , ..., Gm ) .
We give now some applications of this theorem.

a21 b21
 a22 b22 

1). Applying the theorem to the matrix A = 
 ... ... , and then using the triangle inequality
a2m b2m
v
u m
m
m
m
m
m
X
X
u X X
X
X
2
2
t
|xk | ≥
xk yields
ak
bk ≥
|ak bk | ≥
ak bk . Squaring both sides

k=1
k=1
k=1
k=1
k=1
k=1
gives the Cauchy-Schwarz inequalitiy.
2). Let c1 , c2 , ..., cm be m positive rational numbers with the sum 1. Let M be the smallest
Pm common
dk
denominator of c1 , c2 , ..., cm and put ck = M , for k = 1, 2, ..., m. Then, of course,
k=1 dk = M.
Suppose we have m sequences of positive real numbers: a11 , a12 , ..., a1n , a21 , a22 , ..., a2n ,
..., am1 , am2 , ..., amn . Consider the matrix A with n rows and d1 +
d2 + ... + dm columns, where
a
,
a
,
...,
a
each of the first d1 columns equals the
sequence
11 12
1n , each of the next d2 columns
equals the sequence a21 , a22 , ..., a2n , and so on.
Applying now the theorem 4 to the matrix A we get the (generalized) Hölder’s inequality:
n
X
k=1
a1k
n
c1 X
k=1
n
n
c2 X
cm X
m
a2k ...
amk
≥
ac1k1 ac2k2 ...acmk
.
k=1
k=1
3). Suppose now that n, k and m are positive integers, k ≤ m. Let a1 , an ,..., an be nonnegative real
50
numbers. Consider the matrix A with n rows and m columns:
 m m
a1 a1 ... am
1 1 ...
1
m
m
 am
a2 ... a2 1 1 ...
2
A=
 ... ... ... ... ... ... ...
am
am
... am
1 1 ...
n
n
n

1
1 
,
... 
1
where the first k columns are identical and the elements in the remaining m − k columns are only
1’s.
The theorem 4 applied to this matrix yields the Power Mean inequality
k 1
m
1
m
m
m
k
k
k
(a + a2 + · · · an ) ≥
(a + a2 + · · · an )
n 1
n 1
4). Several exercises from the list of problems can be directly solved
by the use ofthe theorem 4,
1 1 ... 1
for a suitable choice of the matrix A. For example, taking A =
and applying
x1 x2 ... xn
the theorem gives an immediate solution to problems 20 and 36.
RECENT PROBLEMS
Problem 71. (IMO 2005) Let x, y, z be positive real numbers such that xyz ≥ 1. Prove that
y5 − y2
z5 − z2
x5 − x2
+
+
≥ 0.
x5 + y 2 + z 2 x2 + y 5 + z 2 x2 + y 2 + z 5
Solution. This problem was considered by the jury of the IMO 2005 as a hard and was the
third problem on the exam. The standard solution is rather long. One may reduce the problem to
the case where xyz = 1 only, then get rid of the denominators, homogenize and finally make use
of the Muirhead’s inequality. The procedure is rather standard although the calculations are quite
tedious. However, during the competition one of the students came up with the following, very
short and extremely elegant solution.
The annoying part of the left hand side is that the denominators have terms with different powers.
In order to get rid of this obstacle one may consider the following inequality:
x5 − x2
x2 (x3 − 1)
≥
.
x5 + y 2 + z 2
x5 + (y 2 + z 2 )x3
This inequality is obviously true if x ≥ 1. It remains true even for 0 < x < 1, since in this case
the nominators are negative. Hence
x2 − x1
x5 − x2
≥ 2
.
x5 + y 2 + z 2
x + y2 + z2
51
Applying the same argument to the other two terms of the left-hand side of the given inequality,
we find out that
x5 − x2
1
1
1
y5 − y2
z5 − z2
1
2
2
2
x
−
+
y
−
+
z
−
.
+
+
≥
x5 + y 2 + z 2 x2 + y 5 + z 2 x2 + y 2 + z 5
x2 + y 2 + z 2
x
y
z
1
1
1
1 1 1
What now remains to show is that x2 − +y 2 − +z 2 − ≥ 0, i.e. that x2 +y 2 +z 2 ≥ + + ,
x
y
z
x y z
for positive reals x, y, z with xyz ≥ 1. This however is a very simple task.
52