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Jensen's Inequality

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Jensen's Inequality is an inequality discovered by a mathematician of that name in 1906.

Contents

Inequality

Let {F} be a convex function of one real variable. Let x_1,\dots,x_n\in\mathbb R and let a_1,\dots, a_n\ge 0 satisfy a_1+\dots+a_n=1. Then


F(a_1x_1+\dots+a_n x_n)\le a_1F(x_1)+\dots+a_n F(x_n)


Proof

The proof of Jensen's inequality is very simple. Since the graph of every convex function lies above its tangent line at every point, we can compare the function {F} with the linear function {L}, whose graph is tangent to the graph of {F} at the point a_1x_1+\dots+a_n x_n. Then the left hand side of the inequality is the same for {F} and {L}, while the right hand side is smaller for {L}. But the inequality for {L} is an identity!

The simplest example of the use of Jensen's inequality is the quadratic mean - arithmetic mean inequality. Take F(x)=x^2 (verify that F'(x)=2x and F''(x)=2>0) and a_1=\dots=a_n=\frac 1n. You'll get \left(\frac{x_1+\dots+x_n}{n}\right)^2\le \frac{x_1^2+\dots+ x_n^2}{n}. Similarly, arithmetic mean-geometric mean inequality can be obtained from Jensen's inequality by considering F(x)=-\log x.

Problems

Introductory

Seeing as this is quite a complicated theorem, there are no introductory problems.

Intermediate

  • Prove that for any \triangle ABC, we have \sin{A}+\sin{B}+\sin{C}\leq \frac{3\sqrt{3}}{2}.

Olympiad

  • Let a,b,c be positive real numbers. Prove that

\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1 (Source)

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