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
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algebra-precalculus
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0I edited your post, see if that's what you wanted to write. – 2012-12-07
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1*I want to be direct, I need to know what you tried, can you indicate it for me? PLEASE.* – 2012-12-07
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0Note 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) – 2012-12-07
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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