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Difference between revisions of "Jensen's Inequality"

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'''Jensen's Inequality''' is an inequality discovered by a mathematician of that name in 1906.
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'''Jensen's Inequality''' is an inequality discovered by Danish mathematician Johan Jensen in 1906.
 
==Inequality==
 
==Inequality==
 
Let <math>{F}</math> be a [[convex function]] of one real variable. Let <math>x_1,\dots,x_n\in\mathbb R</math> and let <math>a_1,\dots, a_n\ge 0</math> satisfy <math>a_1+\dots+a_n=1</math>. Then
 
Let <math>{F}</math> be a [[convex function]] of one real variable. Let <math>x_1,\dots,x_n\in\mathbb R</math> and let <math>a_1,\dots, a_n\ge 0</math> satisfy <math>a_1+\dots+a_n=1</math>. Then
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<math>F(a_1x_1+\dots+a_n x_n)\le a_1F(x_1)+\dots+a_n F(x_n)</math>
 
<math>F(a_1x_1+\dots+a_n x_n)\le a_1F(x_1)+\dots+a_n F(x_n)</math>
 
</center><br>
 
</center><br>
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If <math>{F}</math> is a [[Concave Function]], we have:
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If <math>{F}</math> is a concave function, we have:
 
<br><center>
 
<br><center>
 
<math>F(a_1x_1+\dots+a_n x_n)\ge a_1F(x_1)+\dots+a_n F(x_n)</math>
 
<math>F(a_1x_1+\dots+a_n x_n)\ge a_1F(x_1)+\dots+a_n F(x_n)</math>
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==Proof==
 
==Proof==
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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)
 
  
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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>.
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We only prove the case where <math>F</math> is concave. The proof for the other case is similar.
 +
 
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Let <math>\bar{x}=\sum_{i=1}^n a_ix_i</math>.
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As <math>F</math> is concave, its derivative <math>F'</math> is monotonically decreasing. We consider two cases.
 +
 
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If <math>x_i \le \bar{x}</math>, then
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<cmath>\int_{x_i}^{\bar{x}} F'(t) \, dt \ge \int_{x_i}^{\bar{x}} F'(\bar{x}) \, dt .</cmath>
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If <math>x_i > \bar{x}</math>, then
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<cmath>\int_{\bar{x}}^{x_i} F'(t) \, dt \le \int_{\bar{x}}^{x_i} F'(\bar{x}) \, dt .</cmath>
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By the fundamental theorem of calculus, we have
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<cmath>\int_{x_i}^{\bar{x}} F'(t) \, dt = F(\bar{x}) - F(x_i) .</cmath>
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Evaluating the integrals, each of the last two inequalities implies the same result:
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<cmath>F(\bar{x})-F(x_i) \ge F'(\bar{x})(\bar{x}-x_i)</cmath>
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so this is true for all <math>x_i</math>. Then we have
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<cmath>
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\begin{align*}
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&& F(\bar{x})-F(x_i) &\ge F'(\bar{x})(\bar{x}-x_i) \\
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\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 \\
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\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 \\
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\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
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\end{align*}
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</cmath>
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as desired.
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 +
==Example==
 +
One of the simplest examples of Jensen's inequality is the [[quadratic mean]] - [[arithmetic mean]] inequality. Taking <math>F(x)=x^2</math>, which is convex (because <math>F'(x)=2x</math> and <math>F''(x)=2>0</math>), and <math>a_1=\dots=a_n=\frac 1n</math>, we obtain
 +
<cmath>\left(\frac{x_1+\dots+x_n}{n}\right)^2\le \frac{x_1^2+\dots+ x_n^2}{n} .</cmath>
 +
 
 +
Similarly, [[arithmetic mean]]-[[geometric mean]] inequality ([[AM-GM]]) can be obtained from Jensen's inequality by considering <math>F(x)=-\log x</math>.
 +
 
 +
In fact, the [[power mean inequality]], a generalization of AM-GM, follows from Jensen's inequality.
  
 
==Problems==
 
==Problems==
 +
 
===Introductory===
 
===Introductory===
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Seeing as this is quite a complicated theorem, there are no introductory problems.
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 +
====Problem 1====
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Prove AM-GM using Jensen's Inequality
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 +
====Problem 2====
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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===
 
===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>.
 
* Prove that for any <math>\triangle ABC</math>, we have <math>\sin{A}+\sin{B}+\sin{C}\leq \frac{3\sqrt{3}}{2}</math>.
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* Show that in any triangle <math>\triangle ABC</math> we have <math>\cos {A} \cos{B} \cos {C} \leq \frac{1}{8}</math>
  
 
===Olympiad===
 
===Olympiad===
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<math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1</math> ([[2001 IMO Problems/Problem 2|Source]])
 
<math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1</math> ([[2001 IMO Problems/Problem 2|Source]])
  
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[[Category:Inequality]]
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[[Category:Algebra]]
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[[Category:Theorems]]
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[[Category:Inequalities]]

Latest revision as of 12:10, 20 February 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*} as desired.

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)

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