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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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==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
 
<br><center>
 
<br><center>
 
<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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==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 <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 inequality for <math>{L}</math> is an identity!
 
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 inequality for <math>{L}</math> is an identity!
  
 
The simplest example of the use 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>.
 
The simplest example of the use 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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==Problems==
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===Introductory===
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Seeing as this is quite a complicated theorem, there are no introductory problems.
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===Intermediate===
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===Olympiad===
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*Let <math>a,b,c</math> be positive real numbers. Prove that
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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]])
  
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{{wikify}}
 
 
[[Category:Inequality]]
 
[[Category:Inequality]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Revision as of 19:44, 23 November 2007

Jensen's Inequality is an inequality discovered by a mathematician of that name 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


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


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

Problems

Introductory

Seeing as this is quite a complicated theorem, there are no introductory problems.

Intermediate

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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