Learn/Algebra/Jensen's Inequality

Algebra • Topic 35

Jensen's Inequality

Jensen's Inequality relates the value of a convex function at an average to the average of the function values. It is the theoretical basis for AM-GM and many other inequalities.

Convex Functions

Definition (Convexity)

A function

f

is convex (concave up) on an interval

I

if for all

x, y \in I

and

t \in [0, 1]

f (t x + (1 - t) y) \leq t f (x) + (1 - t) f (y)

Geometrically, the chord connecting any two points lies above the graph.

Calculus Check: If $f^{''} (x) \geq 0$ for all $x \in I$ , then $f$ is convex. (If $f^{''} (x) \leq 0$ , $f$ is concave).

Statement

Theorem (Jensen's Inequality)

f

is convex on an interval containing

x_{1}, \dots, x_{n}

, and

w_{1}, \dots, w_{n}

are positive weights summing to 1:

f (i = 1 \sum n w_{i} x_{i}) \leq i = 1 \sum n w_{i} f (x_{i})

"The function of the average is less than or equal to the average of the function."

Corollary (Concave Case)

f

is concave (e.g.,

ln x, x

), the inequality is reversed:

f (\sum w_{i} x_{i}) \geq \sum w_{i} f (x_{i})

Applications

1. Proving AM-GM

Let

f (x) = - ln x

. Since

f^{''} (x) = 1/ x^{2} > 0

f

is convex.

- ln (\frac{x _{1} + \dots + x _{n}}{n}) \leq \frac{- ln x _{1} - \dots - ln x _{n}}{n}

Multiplying by

- 1

reverses the inequality:

ln (A M) \geq \frac{1}{n} ln (x_{1} \dots x_{n}) = ln ((x_{1} \dots x_{n})^{1/ n}) = ln (GM)

Exponentiating gives

A M \geq GM

Practice Problems

Exercise (Problem 1)

Use Jensen's Inequality with

f (x) = x^{2}

to prove Cauchy-Schwarz (in the form

(\sum x_{i})^{2} \leq n \sum x_{i}^{2}

Exercise (Problem 2)

Prove that for

△ A BC

sin A + sin B + sin C \leq \frac{3 3}{2}

. (Hint:

sin x

is concave on

[0, π]

Exercise (Problem 3)

Prove that

a^{a} b^{b} c^{c} \geq (ab c)^{(a + b + c) /3}

for

a, b, c > 0