Difference between revisions of "Jensen's Inequality"
Durianaops (talk | contribs) (→Proof) |
Durianaops (talk | contribs) m (→Example) |
||
(6 intermediate revisions by the same user not shown) | |||
Line 5: | Line 5: | ||
<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> | ||
− | If <math>{F}</math> is a | + | 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> | ||
Line 15: | Line 15: | ||
Let <math>\bar{x}=\sum_{i=1}^n a_ix_i</math>. | Let <math>\bar{x}=\sum_{i=1}^n a_ix_i</math>. | ||
− | As <math>F</math> is concave, | + | As <math>F</math> is concave, its derivative <math>F'</math> is monotonically decreasing. We consider two cases. |
If <math>x_i \le \bar{x}</math>, then | If <math>x_i \le \bar{x}</math>, then | ||
Line 23: | Line 23: | ||
By the fundamental theorem of calculus, we have | By the fundamental theorem of calculus, we have | ||
<cmath>\int_{x_i}^{\bar{x}} F'(t) \, dt = F(\bar{x}) - F(x_i) .</cmath> | <cmath>\int_{x_i}^{\bar{x}} F'(t) \, dt = F(\bar{x}) - F(x_i) .</cmath> | ||
− | Evaluating the integrals, the last two inequalities | + | Evaluating the integrals, each of the last two inequalities implies the same result: |
+ | <cmath>F(\bar{x})-F(x_i) \ge F'(\bar{x})(\bar{x}-x_i)</cmath> | ||
+ | so this is true for all <math>x_i</math>. Then we have | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
Line 34: | Line 36: | ||
as desired. | as desired. | ||
− | One of the simplest examples of Jensen's inequality is the [[quadratic mean]] - [[arithmetic mean]] inequality. | + | ==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== |
Revision as of 09:44, 31 July 2020
Jensen's Inequality is an inequality discovered by Danish mathematician Johan Jensen in 1906.
Inequality
Let be a convex function of one real variable. Let and let satisfy . Then
If is a concave function, we have:
Proof
We only prove the case where is concave. The proof for the other case is similar.
Let . As is concave, its derivative is monotonically decreasing. We consider two cases.
If , then If , then By the fundamental theorem of calculus, we have Evaluating the integrals, each of the last two inequalities implies the same result: so this is true for all . Then we have as desired.
Example
One of the simplest examples of Jensen's inequality is the quadratic mean - arithmetic mean inequality. Taking , which is convex (because and ), and , we obtain
Similarly, arithmetic mean-geometric mean inequality (AM-GM) can be obtained from Jensen's inequality by considering .
In fact, the power mean inequality, a generalization of AM-GM, follows from Jensen's inequality.
Problems
Introductory
Prove AM-GM using Jensen's Inequality
Intermediate
- Prove that for any , we have .
- Show that in any triangle we have
Olympiad
- Let be positive real numbers. Prove that
(Source)