Difference between revisions of "Jensen's Inequality"

Latest revision as of 05:29, 28 November 2024

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

We only prove the case where $F$ is concave. The proof for the other case is similar.

Let $\bar{x}=\sum_{i=1}^n a_ix_i$ . As $F$ is concave, its derivative $F'$ is monotonically decreasing. We consider two cases.

If $x_i \le \bar{x}$ , then $\int_{x_i}^{\bar{x}} F'(t) \, dt \ge \int_{x_i}^{\bar{x}} F'(\bar{x}) \, dt .$ If $x_i > \bar{x}$ , then $\int_{\bar{x}}^{x_i} F'(t) \, dt \le \int_{\bar{x}}^{x_i} F'(\bar{x}) \, dt .$ By the fundamental theorem of calculus, we have $\int_{x_i}^{\bar{x}} F'(t) \, dt = F(\bar{x}) - F(x_i) .$ Evaluating the integrals, each of the last two inequalities implies the same result: $F(\bar{x})-F(x_i) \ge F'(\bar{x})(\bar{x}-x_i)$ so this is true for all $x_i$ . Then we have $\begin{align*} && F(\bar{x})-F(x_i) &\ge F'(\bar{x})(\bar{x}-x_i) \\ \Longrightarrow && a_i F(\bar{x}) - a_i F(x_i) &\ge F'(\bar{x})(a_i\bar{x}-a_i x_i) && \text{as } a_i>0 \\ \Longrightarrow && F(\bar{x}) - \sum_{i=1}^n a_i F(x_i) &\ge F'(\bar{x})\left(\bar{x} - \sum_{i=1}^n a_i x_i \right) && \text{as } \sum_{i=1}^n a_i = 1 \\ \Longrightarrow && F(\bar{x}) &\ge \sum_{i=1}^n a_i F(x_i) && \text{as } \bar{x}=\sum_{i=1}^n a_ix_i \end{align*}$ which is exactly what we want! Hooray!😀😀😀

Example

One of the simplest examples of Jensen's inequality is the quadratic mean - arithmetic mean inequality. Taking $F(x)=x^2$ , which is convex (because $F'(x)=2x$ and $F''(x)=2>0$ ), and $a_1=\dots=a_n=\frac 1n$ , we obtain $\left(\frac{x_1+\dots+x_n}{n}\right)^2\le \frac{x_1^2+\dots+ x_n^2}{n} .$

Similarly, arithmetic mean-geometric mean inequality (AM-GM) can be obtained from Jensen's inequality by considering $F(x)=-\log x$ .

In fact, the power mean inequality, a generalization of AM-GM, follows from Jensen's inequality.

Problems

Introductory

Problem 1

Prove AM-GM using Jensen's Inequality

Problem 2

Prove the weighted AM-GM inequality. (It states that $x_1^{\lambda_1}x_2^{\lambda_2} \cdots x_n^{\lambda_n} \leq \lambda_1 x_1 + \lambda_2 x_2 +\cdots+ \lambda_n x_n$ when $\lambda_1 + \cdots + \lambda_n = 1$ )

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 34: / Line 34: @@
 \end{align*}
 </cmath>
-as desired.
+which is exactly what we want! Hooray!😀😀😀
 ==Example==
@@ Line 52: / Line 52: @@
 ====Problem 2====
-Prove that <math>x_1^{\lambda_1}x_2^{\lambda_2} \cdots x_n^{\lambda_n} \leq \lambda_1 x_1 + \lambda_2 x_2 +\cdots+ \lambda_n x_n</math> when <math>\lambda_1 + \cdots + \lambda_n = 1</math>
+Prove the weighted [[AM-GM Inequality|AM-GM inequality]]. (It states that <math>x_1^{\lambda_1}x_2^{\lambda_2} \cdots x_n^{\lambda_n} \leq \lambda_1 x_1 + \lambda_2 x_2 +\cdots+ \lambda_n x_n</math> when <math>\lambda_1 + \cdots + \lambda_n = 1</math>)
 ===Intermediate===

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