PUTNAM TRAINING
INEQUALITIES
(Last updated: November 6, 2012)
Remark. This is a list of exercises on inequalities. —Miguel A. Lerma
Exercises
1. If a, b, c > 0, prove that (a2 b + b2 c + c2 a)(ab2 + bc2 + ca2 ) ≥ 9a2 b2 c2 .
n
n+1
2. Prove that n! <
, for n = 2, 3, 4, . . . ,
2
3. If 0 < p, 0 < q, and p + q < 1, prove that (px + qy)2 ≤ px2 + qy 2 .
p
√
√
√
4. If a, b, c ≥ 0, prove that 3(a + b + c) ≥ a + b + c.
5. Let x, y, z > 0 with xyz = 1. Prove that x + y + z ≤ x2 + y 2 + z 2 .
6. Show that
q
q
p
a21 + b21 + a22 + b22 + · · · + a2n + b2n ≥
p
(a1 + a2 + · · · + an )2 + (b1 + b2 + · · · + bn )2
7. Find the minimum value of the function f (x1 , x2 , . . . , xn ) = x1 + x2 + · · · + xn , where
x1 , x2 , . . . , xn are positive real numbers such that x1 x2 · · · xn = 1.
8. Let x, y, z ≥ 0 with xyz = 1. Find the minimum of
S=
x2
y2
z2
+
+
.
y+z z+x x+y
9. If x, y, z > 0, and x + y + z = 1, find the minimum value of
1 1 1
+ + .
x y z
10. Prove that in a triangle with sides a, b, c and opposite angles A, B, C (in radians)
the following relation holds:
aA + bB + cC
π
≥ .
a+b+c
3
1
PUTNAM TRAINING
INEQUALITIES
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11. (Putnam, 2003) Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn nonnegative real numbers. Show
that
(a1 a2 · · · an )1/n + (b1 b2 · · · bn )1/n ≤ ((a1 + b1 )(a2 + b2 ) · · · (an + bn ))1/n
12. The notation n!(k) means take factorial of n k times. For example, n!(3) means ((n!)!)!
What is bigger, 1999!(2000) or 2000!(1999) ?
13. Which is larger, 19991999 or 20001998 ?
14. Prove that there are no positive integers a, b such that b2 + b + 1 = a2 .
15. (Inspired in Putnam 1968, B6) Prove that a polynomial with only real roots and all
coefficients equal to ±1 has degree at most 3.
16. (Putnam 1984) Find the minimum value of
2
√
9
2
(u − v) +
2 − u2 −
v
√
for 0 < u < 2 and v > 0.
1
1
3
2n − 1
1
17. Show that √ ≤
···
<√ .
2
4
2n
4n
2n
18. (Putnam, 2004) Let m and n be positive integers. Show that
(m + n)!
m! n!
<
.
(m + n)m+n
mm nn
19. Let a1 , a2 , . . . , an be a sequence of positive numbers, and let b1 , b2 , . . . , bn be any
permutation of the first sequence. Show that
a1 a2
an
+
+ ··· +
≥ n.
b1
b2
bn
20. (Rearrangement Inequality.) Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn increasing sequences
of real numbers, and let x1 , x2 , . . . , xn be any permutation of b1 , b2 , . . . , bn . Show that
n
X
i=1
ai bi ≥
n
X
ai xi .
i=1
21. Prove that the p-mean tends to the geometric mean as p approaches zero. In other
other words, if a1 , . . . , an are positive real numbers, then
!1/p
!1/n
n
n
Y
1X p
=
ak
lim
a
p→0
n k=1 k
k=1
PUTNAM TRAINING
INEQUALITIES
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22. If a, b, and c are the sides of a triangle, prove that
b
c
a
+
+
≥ 3.
b+c−a c+a−b a+b−c
23. Here we use Knuth’s up-arrow notation: a ↑ b = ab , a ↑↑ b = a ↑ (a ↑ (. . . ↑ a)), so
|
{z
}
b copies of a
22
e.g. 2 ↑↑ 3 = 2 ↑ (2 ↑ 2)) = 2 . What is larger, 2 ↑↑ 2011 or 3 ↑↑ 2010?
24. Prove that e1/e + e1/π ≥ 2e1/3 .
25. Prove that the function f (x) =
a1 + · · · + an
.
n
n
X
(x − ai )2 attains its minimum value at x = a =
i=1
26. Find the positive solutions of the system of equations
1
1
1
x1 +
= 4 , x2 +
= 1 , . . . , x99 +
= 4,
x2
x3
x100
x100 +
1
= 1.
x1
27. Prove that if the numbers a, b, and c satisfy the inequalities |a − b| ≥ |c|, |b − c| ≥ |a|,
|c − a| ≥ |b|, then one of those numbers is the sum of the other two.
28. Find the minimum of sin3 x/ cos x + cos3 x/ sin x, 0 < x < π/2.
29. Let ai > 0, i = 1, . . . , n, and s = a1 + · · · + an . Prove
a1
a2
an
n
+
+ ··· +
≥
.
s − a1 s − a2
s − an
n−1
PUTNAM TRAINING
INEQUALITIES
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Hints
1. One way to solve this problem is by using the Arithmetic Mean-Geometric Mean
inequality on each factor of the left hand side.
2. Apply the Arithmetic Mean-Geometric Mean inequality to the set of numbers 1, 2, . . . , n.
3. Power means inequality with weights
p
p+q
and
q
.
p+q
4. Power means inequality.
5. —
6. This problem can be solved by using Minkowski’s inequality, but another way to look
at it is by an appropriate geometrical interpretation of the terms (as distances between
points of the plane.)
7. Many minimization or maximization problems are inequalities in disguise. The solution usually consists of “guessing” the maximum or minimum value of the function, and then proving that it is in fact maximum or minimum. In this case, given
the symmetry of the function a good guess is f (1, 1, . . . , 1) = n, so try to prove
f (x1 , x2 , . . . , xn ) ≥ n. Use the Arithmetic Mean-Geometric Mean inequality on
x1 , . . . , x n .
y
x
z
8. Apply the Cauchy-Schwarz inequality to the vectors ( √y+z
, √x+y
, √z+x
) and (u, v, w),
and choose appropriate values for u, v, w.
9. Arithmetic-Harmonic Mean inequality.
10. Assume a ≤ b ≤ c, A ≤ B ≤ C, and use Chebyshev’s Inequality.
11. Divide by the right hand side and use the Arithmetic Mean-Geometric Mean inequality
on both terms of the left.
12. Note that n! is increasing (n < m =⇒ n! < m!)
13. Look at the function f (x) = (1999 − x) ln (1999 + x).
14. The numbers b2 and (b + 1)2 are consecutive squares.
15. Use the Arithmetic Mean-Geometric Mean inequality on the squares of the roots of
the polynomial.
16. Think geometrically. Interpret the given expression as the square of the distance
between two points in the plane. The problem becomes that of finding the minimum
distance between two curves.
PUTNAM TRAINING
17. Consider the expressions P =
k
< k+1
, for k = 1, 2, . . . .
that k−1
k
INEQUALITIES
1
2
3
4
···
2n−1
2n
and Q =
2
3
5
4
5
···
2n−2
2n−1
. Note
18. Look at the binomial expansion of (m + n)m+n .
19. Arithmetic Mean-Geometric Mean inequality.
20. Try first the cases n = 1 and n = 2. Then use induction.
21. Take logarithms and use L’Hôpital.
22. Set x = b + c − a, y = c + a − b, z = a + b − c.
2
23. We have 22 = 16 < 27 = 33 .
24. Show that f (x) = e1/x for x > 0 is decreasing and convex.
25. Prove that f (x) − f (a) ≥ 0.
26. By the AM-GM inequality we have x1 +
inequalities are actually equalities.
27. Square both sides of those inequalities.
28. Rearrangement inequality.
29. Rearrangement inequality.
1
x2
q
≥ 2 xx12 , . . . Try to prove that those
PUTNAM TRAINING
INEQUALITIES
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Main Inequalities
1. Arithmetic Mean, Geometric Mean, Harmonic Mean Inequalities. Let a1 , . . . , an
be positive numbers. Then the following inequalities hold:
n
1
1
+ ··· +
a1
an
|
{z
}
≤
√
n
a1 · a2 · · · · · an ≤
|
Harmonic Mean
{z
}
a1 + · · · + an
n
|
Geometric Mean
{z
}
Arithmetic Mean
In all cases equality holds if and only if a1 = · · · = an .
2. Power Means Inequality. The AM-GM, GM-HM and AM-HM inequalities are particular cases of a more general kind of inequality called Power Means Inequality.
Let r be a non-zero real number. We define the r-mean or rth power mean of positive
numbers a1 , . . . , an as follows:
!1/r
n
X
1
ar
.
M r (a1 , . . . , an ) =
n i=1 i
The ordinary arithmetic mean is M 1 , M 2 is the quadratic mean, M −1 is the harmonic mean.
Furthermore we define the 0-mean to be equal to the geometric mean:
n
1/n
Y
ai
.
M (a1 , . . . , an ) =
0
i=1
Then, for any real numbers r, s such that r < s, the following inequality holds:
M r (a1 , . . . , an ) ≤ M s (a1 , . . . , an ) .
Equality holds if and only if a1 = · · · = an .
2.1. Power Means Sub/Superadditivity. Let a1 , . . . , an , b1 , . . . , bn be positive real numbers.
(a) If r > 1, then the r-mean is subadditive, i.e.:
M r (a1 + b1 , . . . , an + bn ) ≤ M r (a1 , . . . , an ) + M r (b1 , . . . , bn ) .
(b) If r < 1, then the r-mean is superadditive, i.e.:
M r (a1 + b1 , . . . , an + bn ) ≥ M r (a1 , . . . , an ) + M r (b1 , . . . , bn ) .
Equality holds if and only if (a1 , . . . , an ) and (b1 , . . . , bn ) are proportional.
3. Cauchy-Schwarz.
n
X
i=1
ai bi
2
≤
n
X
i=1
a2i
n
X
2
bi .
i=1
PUTNAM TRAINING
INEQUALITIES
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4. Hölder. If p > 1 and 1/p + 1/q = 1 then
n
n
n
X
X
1/p X
1/q
p
p
ai bi ≤
|ai |
|bi |
.
i=1
i=1
i=1
(For p = q = 2 we get Cauchy-Schwarz.)
5. Minkowski. If p > 1 then
n
X
p
|ai + bi |
1/p
≤
i=1
n
X
p
|ai |
1/p
+
n
X
i=1
|bi |p
1/p
.
i=1
Equality holds iff (a1 , . . . , an ) and (b1 , . . . , bn ) are proportional.
6. Norm Monotonicity. If ai > 0 (i = 1, . . . , n), s > t > 0, then
n
X
i=1
asi
1/s
≤
n
1/t
X
ati
.
i=1
7. Chebyshev. Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be sequences of real numbers which are
monotonic in the same direction (we have a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn , or we
could reverse all inequalities.) Then
n
n
n
1 X
1 X
1X
ai bi ≥
ai
bi .
n i=1
n i=1
n i=1
8. Rearrangement Inequality. For every choice of real numbers x1 ≤ · · · ≤ xn and
y1 ≤ · · · ≤ yn , and any permutation xσ(1) , . . . , xσ(n) of x1 , . . . , xn , we have
xn y1 + · · · + x1 yn ≤ xσ(1) y1 + · · · + xσ(n) yn ≤ x1 y1 + · · · + xn yn .
If the numbers are different, e.g., x1 < · · · < xn and y1 < · · · < yn , then the lower bound is
attained only for the permutation which reverses the order, i.e. σ(i) = n − i + 1, and the
upper bound is attained only for the identity, i.e. σ(i) = i, for i = 1, . . . , n.
9. Schur. If x, y, x are positive real numbers and k is a real number such that k ≥ 1, then
xk (x − y)(x − z) + y k (y − x)(y − z) + z k (z − x)(z − y) ≥ 0 .
For k = 1 the inequality becomes
x3 + y 3 + z 3 + 3xyz ≥ xy(x + y) + yz(y + z) + zx(z + x) .
PUTNAM TRAINING
INEQUALITIES
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10. Weighted Power Means Inequality. Let w1 , . . . , wn be positive real numbers such
that w1 + · · · + wn = 1. Let r be a non-zero real number. We define the rth weighted power
mean of non-negative numbers a1 , . . . , an as follows:
!1/r
n
X
Mwr (a1 , . . . , an ) =
wi ari
.
i=1
As r → 0 the rth weighted power mean tends to:
Mw0 (a1 , . . . , an )
=
n
Y
i
aw
.
i
i=1
which we call 0th weighted power mean. If wi = 1/n we get the ordinary rth power means.
Then for any real numbers r, s such that r < s, the following inequality holds:
Mwr (a1 , . . . , an ) ≤ Mws (a1 , . . . , an ) .
11. Convexity. A function f : (a, b) → R is said to be convex if
f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y)
for every x, y ∈ (a, b), 0 ≤ λ ≤ 1. Graphically, the condition is that for x < t < y the point
(t, f (t)) should lie below or on the line connecting the points (x, f (x)) and (y, f (y)).
Figure 1. Convex function.