Jensen's Inequality

Revision as of 21:37, 18 May 2020 by Theepiccarrot7 (talk | contribs) (Intermediate)

Jensen's Inequality is an inequality discovered by Danish mathematician Johan Jensen in 1906.


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:

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


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



Prove AM-GM using Jensen's Inequality


  • 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 $\cos {A} \cos{B} \cos {C} \leq \frac{1}{8}$


  • 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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