Difference between revisions of "Jensen's Inequality"

Revision as of 11:25, 28 May 2010

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

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 equality case holds for all linear functions! (check it yourself)

One of the simplest examples 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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 ==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 <math>{F}</math> with the linear function <math>{L}</math>, whose graph is tangent to the graph of <math>{F}</math> at the point <math>a_1x_1+\dots+a_n x_n</math>. Then the left hand side of the inequality is the same for <math>{F}</math> and <math>{L}</math>, while the right hand side is smaller for <math>{L}</math>. But the inequality for <math>{L}</math> is an identity!
+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 <math>{F}</math> with the linear function <math>{L}</math>, whose graph is tangent to the graph of <math>{F}</math> at the point <math>a_1x_1+\dots+a_n x_n</math>. Then the left hand side of the inequality is the same for <math>{F}</math> and <math>{L}</math>, while the right hand side is smaller for <math>{L}</math>. But the equality case holds for all linear functions!  (check it yourself)
-The simplest example of the use of Jensen's inequality is the [[quadratic mean]] - [[arithmetic mean]] inequality. Take <math>F(x)=x^2</math> (verify that <math>F'(x)=2x</math> and <math>F''(x)=2>0</math>) and <math>a_1=\dots=a_n=\frac 1n</math>. You'll get <math>\left(\frac{x_1+\dots+x_n}{n}\right)^2\le \frac{x_1^2+\dots+ x_n^2}{n} </math>. Similarly, [[arithmetic mean]]-[[geometric mean]] inequality can be obtained from Jensen's inequality by considering <math>F(x)=-\log x</math>.
+One of the simplest examples of Jensen's inequality is the [[quadratic mean]] - [[arithmetic mean]] inequality. Take <math>F(x)=x^2</math> (verify that <math>F'(x)=2x</math> and <math>F''(x)=2>0</math>) and <math>a_1=\dots=a_n=\frac 1n</math>. You'll get <math>\left(\frac{x_1+\dots+x_n}{n}\right)^2\le \frac{x_1^2+\dots+ x_n^2}{n} </math>. Similarly, [[arithmetic mean]]-[[geometric mean]] inequality can be obtained from Jensen's inequality by considering <math>F(x)=-\log x</math>.
 ==Problems==

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