proof of arithmetic-geometric means inequality
A short geometrical proof can be given for the case of the arithmetic-geometric means inequality.
Let and be two non negative numbers. Draw the line such that has length , and has length , as in the following picture, and draw a semicircle with diameter . Let be the center of the circle.
Now raise perpendiculars and to . Notice that is a radius, and so
Also notice that for any point , and equality is obtained only when , that is, when .
Notice also that is a height over the hypotenuse on right triangle . We have then triangle similarities , and thus
which implies . Since , we conclude
This special case can also be proved using rearrangement inequality. Let non negative numbers, and assume . Let , and then . Now suppose and are such that one of them is and the other is . Rearrangement inequality states that is maximum when and . So, we have
and substituting back gives
where it follows the desired result.
One more proof can be given as follows. Let . Then , and equality holds only when . Then, becomes
and substituting back gives the desired result as in the previous proof.
Title | proof of arithmetic-geometric means inequality |
---|---|
Canonical name | ProofOfArithmeticgeometricMeansInequality |
Date of creation | 2013-03-22 14:49:14 |
Last modified on | 2013-03-22 14:49:14 |
Owner | mathcam (2727) |
Last modified by | mathcam (2727) |
Numerical id | 10 |
Author | mathcam (2727) |
Entry type | Proof |
Classification | msc 26D15 |