Difference between revisions of "Jensen's Inequality"
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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> | ||
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− | 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> |
Revision as of 09:33, 31 July 2020
Jensen's Inequality is an inequality discovered by Danish mathematician Johan Jensen in 1906.
Contents
[hide]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. 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)