Let there be $u=( \sqrt{a},\sqrt{b})$ and $v= (\sqrt{b},\sqrt{a})$ where $a,b\in \mathbb{R}$. Using the Schwarz inequality, prove that the geometric mean $\sqrt{ab}$ is not bigger than the arithmetic mean $(a+b)/2$ of them.
Using the Schwarz inequality to prove the geometric mean is not larger than the arithmetic mean
0
$\begingroup$
algebra-precalculus
-
0The person posing the problem has told you exactly how to do it. Take the Cauchy-Schwartz Inequality, case $n=2$, and just substitute the suggested values. – 2012-12-07
1 Answers
2
Given $u=(\sqrt{a},\sqrt{b})$ and $ u=(\sqrt{b},\sqrt{a}) $, you want to prove $ \sqrt{ab}\leq \frac{a+b}{2}.$ Recalling the Cauchy-Schwarz inequality
$ |u.v|\leq ||u||||v||. $
Compute $u.v$, $||u||$, and $||v||$ as
$ u.v = (\sqrt{a},\sqrt{b}).(\sqrt{b},\sqrt{a}) =\sqrt{a}\sqrt{b}+\sqrt{b}\sqrt{a}=2\sqrt{ab},$
$ ||u||= \sqrt{ a + b }, \quad ||v||=\sqrt{ a + b }. $
Now, substitute what we just computed in the Cauchy-Schwarz inequality
$ 2\sqrt{ab} \leq \sqrt{ a + b } \sqrt{ a + b }\implies \sqrt{ab} \leq \frac{(a+b)}{2}. $