proof of arithmetic-geometric means inequality
A short geometrical proof can be given for the case of the arithmetic-geometric means inequality.
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 .
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|
|Date of creation||2013-03-22 14:49:14|
|Last modified on||2013-03-22 14:49:14|
|Last modified by||mathcam (2727)|