Difference between revisions of "Jensen's Inequality"
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==Proof== | ==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 | + | 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 equality case holds for all linear functions! (check it yourself) |
− | + | 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>. | |
==Problems== | ==Problems== |
Revision as of 12:25, 28 May 2010
Jensen's Inequality is an inequality discovered by a mathematician of that name in 1906.
Inequality
Let be a convex function of one real variable. Let
and let
satisfy
. Then
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 with the linear function
, whose graph is tangent to the graph of
at the point
. Then the left hand side of the inequality is the same for
and
, while the right hand side is smaller for
. But the equality case holds for all linear functions! (check it yourself)
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
Seeing as this is quite a complicated theorem, there are no introductory problems.
Intermediate
- Prove that for any
, we have
.
Olympiad
- Let
be positive real numbers. Prove that
(Source)