Difference between revisions of "Jensen's Inequality"

Revision as of 20:37, 18 May 2020

Jensen's Inequality is an inequality discovered by Danish mathematician Johan Jensen 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

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

If ${F}$ is a Concave Function, we have:

$F(a_1x_1+\dots+a_n x_n)\ge a_1F(x_1)+\dots+a_n F(x_n)$

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

Prove AM-GM using Jensen's Inequality

Intermediate

Prove that for any $\triangle ABC$ , we have $\sin{A}+\sin{B}+\sin{C}\leq \frac{3\sqrt{3}}{2}$ .
Show that in any triangle $\triangle ABC$ we have $\cos {A} \cos{B} \cos {C} \leq \frac{1}{8}$

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)

@@ Line 21: / Line 21: @@
 ===Intermediate===
 * Prove that for any <math>\triangle ABC</math>, we have <math>\sin{A}+\sin{B}+\sin{C}\leq \frac{3\sqrt{3}}{2}</math>.
-* Show that in any triangle <math>\triangle ABC</math> we have <math>\cos A \cosB \cos C \leq \frac{1}{8}</math>
+* Show that in any triangle <math>\triangle ABC</math> we have <math>\cos {A} \cos{B} \cos {C} \leq \frac{1}{8}</math>
 ===Olympiad===

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