Difference between revisions of "Jensen's Inequality"
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==Proof== | ==Proof== | ||
− | The proof | + | |
+ | We only prove the case where <math>F</math> is concave. The proof for the other case is similar. | ||
+ | |||
+ | Let <math>\bar{x}=\sum_{i=1}^n a_ix_i</math>. | ||
+ | As <math>F</math> is concave, then its derivative <math>F'</math> is monotonically decreasing. We consider two cases. | ||
+ | |||
+ | If <math>x_i \le \bar{x}</math>, then | ||
+ | <cmath>\int_{x_i}^{\bar{x}} F'(t) \, dt \ge \int_{x_i}^{\bar{x}} F'(\bar{x}) \, dt .</cmath> | ||
+ | If <math>x_i > \bar{x}</math>, then | ||
+ | <cmath>\int_{\bar{x}}^{x_i} F'(t) \, dt \le \int_{\bar{x}}^{x_i} F'(\bar{x}) \, dt .</cmath> | ||
+ | By the fundamental theorem of calculus, we have | ||
+ | <cmath>\int_{x_i}^{\bar{x}} F'(t) \, dt = F(\bar{x}) - F(x_i) .</cmath> | ||
+ | Evaluating the integrals, the last two inequalities both result in | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | as desired. | ||
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>. | 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>. |
Revision as of 08:21, 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, then its derivative is monotonically decreasing. We consider two cases.
If , then If , then By the fundamental theorem of calculus, we have Evaluating the integrals, the last two inequalities both result in as desired.
One of the simplest examples of Jensen's inequality is the quadratic mean - arithmetic mean inequality. Take (verify that and ) and . You'll get . Similarly, arithmetic mean-geometric mean inequality can be obtained from Jensen's inequality by considering .
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)