Difference between revisions of "Jensen's Inequality"
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+ | '''Jensen's Inequality''' is an inequality discovered by Danish mathematician Johan Jensen in 1906. | ||
+ | ==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> | ||
− | + | If <math>{F}</math> is a concave function, we have: | |
+ | <br><center> | ||
+ | <math>F(a_1x_1+\dots+a_n x_n)\ge a_1F(x_1)+\dots+a_n F(x_n)</math> | ||
+ | </center><br> | ||
+ | |||
+ | ==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, 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, 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> | ||
+ | \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. | ||
+ | |||
+ | ==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== | ||
+ | |||
+ | ===Introductory=== | ||
+ | |||
+ | ====Problem 1==== | ||
+ | Prove AM-GM using Jensen's Inequality | ||
+ | |||
+ | ====Problem 2==== | ||
+ | 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=== | |
+ | * Prove that for any <math>\triangle ABC</math>, we have <math>\sin{A}+\sin{B}+\sin{C}\leq \frac{3\sqrt{3}}{2}</math>. | ||
+ | * Show that in any triangle <math>\triangle ABC</math> we have <math>\cos {A} \cos{B} \cos {C} \leq \frac{1}{8}</math> | ||
− | {{ | + | ===Olympiad=== |
+ | *Let <math>a,b,c</math> be positive real numbers. Prove that | ||
+ | <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]]) | ||
− | + | [[Category:Algebra]] | |
− | [[Category: | + | [[Category:Inequalities]] |
− | [[Category: |
Latest revision as of 13:10, 20 February 2024
Jensen's Inequality is an inequality discovered by Danish mathematician Johan Jensen in 1906.
Contents
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
Problem 1
Prove AM-GM using Jensen's Inequality
Problem 2
Prove the weighted AM-GM inequality. (It states that when
)
Intermediate
- Prove that for any
, we have
.
- Show that in any triangle
we have
Olympiad
- Let
be positive real numbers. Prove that
(Source)