Jensen's 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

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

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 $F(x)=-\log x$ .

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Jensen's Inequality