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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.

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I edited your post, see if that's what you wanted to write. – Mhenni Benghorbal Dec 7 '12 at 6:15
I want to be direct, I need to know what you tried, can you indicate it for me? PLEASE. – Did Dec 7 '12 at 6:19
Note that Schwartz says that $||uv||\leq ||u||^{1/2} ||v||^{1/2}$ which says $||uv||^2\leq ||u|| ||v||$. If you recall that $||(x,y)||= \sqrt{x^2+y^2}$, can you try unraveling? Note by uv I mean if u=(x,y) and v=(z,w), uv=(xz, yz) – user45150 Dec 7 '12 at 6:23
The 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. – André Nicolas Dec 7 '12 at 6:30

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

$$ 2\sqrt{ab} \leq \sqrt{ a + b } \sqrt{ a + b }\implies \sqrt{ab} \leq \frac{(a+b)}{2}. $$

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