Lecture 21 Inequalities
Another way in which optimization can be applied is to prove inequalities.
Arithmetic Geometric mean inequality Let a and b be positive numbers then
1
(a + b).
2
1
1
a2 b2 ≤
This can be proved in a purely algebraic way.
Algebraic proof of arithmetic geometric mean inequality
1
1
1
1
a + b − 2a 2 b 2 = (a 2 − b 2 )2 .
Analytic proof of arithmetic geometric mean inequality
It suffices to prove the inequality when a + b = 1. This is because
1
1
1
1
(ta) 2 (tb) 2 = ta 2 b 2 ,
while
(ta + tb) = t(a + b),
1
so we just pick t = a+b
.
Thus what we need to prove is
√ √
1
x 1−x≤ ,
2
when 0 < x < 1. We let
f (x) =
and calculate
√ √
x 1 − x,
√
√
1−x
x
√
f (x) =
− √
.
2 x
2 1−x
√
√
x
1−x
1
00
f (x) = − √ √
−
.
3 −
3
2 x 1 − x 4(1 − x) 2
4x 2
0
All terms in the last line are negative so f is strictly concave. The unique critical point is
at x = 21 , where equality holds. We have shown that
√ √
1
x 1−x≤ ,
2
since
1
2
is the maximum.
1
The analytic proof looks a lot messier than the algebraic one, but it is more powerful.
For instance, by the same methods, we get that if α, β > 0 and
α + β = 1,
then
aα bβ ≤ αa + βb,
for a, b > 0.
This simply requires applying concavity for the function
f (x) = xα (1 − x)β ,
and finding the unique maximum.
Closely related to the arithmetic geometric mean inequality us the geometric harmonic
mean inequality. The simplest version is:
Simple version of Harmonic Geometric mean inequality Let a, b > 0 be real
numbers:
1 1
2
2 2
1
1 ≤a b .
a + b
Proof Multiply numerator and denominator of the left hand side by ab. Then divide
1 1
both sides by a 2 b 2 and take the reciprocal and you obtain the arithmetic geometric mean
inequality. All steps are reversible so that the two inequalities are equivalent.
In the same way, we can obtain a more general (weighted) version. Let α, β > 0 with
α + β = 1. Then
1
≤ a α bβ .
β
α
+
a
b
We can obtain similar results for sums of not just two terms but n terms.
n-term AGM inequality Let α1 , . . . , αn > 0 with
n
X
αj = 1.
j=1
Then
1 α2
aα
1 a2
n
. . . aα
n
≤
n
X
αj aj .
j=1
Proof of n-term AGM inequality We prove this by induction on n. The base case,
n = 2 is already known. We let
n−1
X
α=
αj
j=1
2
and β = αn . We let
1
α
n−1
1
α
a = (aα
1 . . . an−1 ) ,
and b = an . Then using the two term AGM inequality, we obtain
αn
1 α2
aα
1 a2 . . . an ≤ αa + βb.
We now simply apply the n − 1 term AGM to αa to obtain the desired result.
Similarly we could write down an n-term harmonic-geometric mean inequality.
Discrete Hölder inequality Let p, q > 0 and p1 + 1q = 1. Let a1 , . . . an , b1 , . . . , bn > 0
be real numbers. Then
n
n
n
X
X
1 X q 1
aj bj ≤ (
apj ) p (
bk ) q .
j=1
j=1
k=1
1
p
Proof of discrete Hölder By AGM with α =
and β = 1q , we get
ap
bq
+ .
p
q
ab ≤
Applying this to each term in the sum, we get
n
n
X
n
1X p 1X q
aj +
bk .
aj bj ≤
p
q
j=1
j=1
k=1
Unfortunately, the right hand side is always larger than what we want by AGM. However,
Hölder’s inequality doesn’t change if we multiply all a’s by a given positive
all
Pn constantPand
n
b’s by a given positive constant. So we may restrict to the case that j=1 apj and k=1 bqk
are both equal to 1. In that case the right hand side is exactly 1, which is what we want.
We can also obtain an integral version.
Hölder’s inequality Let p, q > 0 and
functions on an interval [a, b]. Then
Z
b
Z
f (x)g(x)dx ≤ (
a
b
1
p
+
1
q
= 1. Let f, g be nonnegative integrable
Z
f (x) dx) (
p
a
1
p
b
1
g(x)q dx) q .
a
To prove this, we just apply the discrete Hölder’s inequality to Riemann sums.
We can apply Hölder’s inequality to estimate means. To wit with f nonnegative and
integrable and p, q as above:
3
1
b−a
b
Z
f (x)dx
a
b
Z
1
=
b−a
f (x) · 1dx
a
1
≤(
b−a
Z
1
b−a
Z
=(
b
1
1 dx) (
b−a
q
a
1
q
b
b
Z
1
f (x)p dx) p
a
1
f (x)p dx) p .
a
This inequality says that the pth root of the mean pth power of f is greater than or
equal to the mean of f as long as p > 1. A slightly more general formulation is
Jensen’s Inequality Let g be a convex function and f as before then
1
g(
b−a
Z
a
b
1
f (x)dx) ≤
b−a
Z
b
g(f (x))dx.
a
We can build up a proof of this starting from sums of two terms, generalizing to sums
of n terms by induction, and ultimately to integrals by applying the n-term version to
Riemann sums. The two term version:
g(αa + βb) ≤ αg(a) + βg(b),
is self-evidently the definition of convexity of g. Thus we have come full circle. We can
think of Hölder’s inequality as being true because the function xp is convex.
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