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Prove that $\left (\dfrac{a^2 + b^2 +c^2}{a+b+c} \right) ^ {(a+b+c)} > a^a b^b c^c$ if $a$, $b$ and $c$ are distinct natural numbers. Is it possible using induction?

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up vote 12 down vote accepted

I recommend to read the answer from here. section Weighted AM-GM Inequality :) it is a good one .

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nice link! (+1) – user 1618033 Aug 23 '12 at 19:14
@downvoter why (-1)? – Iuli Feb 5 '13 at 12:56
wasn't me but I guess because the linked page is no longer available – Valentin Feb 10 '13 at 0:46
@Iuli Answers that only include a link are bad, for exactly this reason - very often the page that was linked to disappears. Can you edit your answer so that it's useful again? – Chris Taylor Feb 13 '13 at 12:19

Rewrite it as: $$ \left( a \frac{a}{a+b+c} + b \frac{b}{a+b+c} + c \frac{c}{a+b+c} \right) > a^\frac{a}{a+b+c} \cdot b^\frac{b}{a+b+c} \cdot c^\frac{c}{a+b+c} $$ This is Jensen's inequality: $$ \log\left(\mathsf{E}\left(X\right)\right) > \mathsf{E}\left(\log\left(X\right)\right) \quad \text{or}\quad \mathsf{E}\left(X\right) > \exp \left( \mathsf{E}\left(\log\left(X\right)\right) \right) $$ where $X$ is the random variable which can assume one of three possible values $\{a,b,c\}$ with respective probabilities $\{ \frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c} \}$.

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@Fixee Thanks for the edits. – Sasha Aug 23 '12 at 16:45
Nice solution! (+1) – user 1618033 Aug 23 '12 at 19:15

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