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- #1

- Feb 14, 2012

- 3,909

- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,909

$a^{a} \ b^{b} \dfrac{(a+b)!}{a! \ b!} \leq (a+b)^{a+b}$

This inequality can be expressed as

${{a+b}\choose{b}} \ a^{a} \ b^{b} \leq \sum_{k=0}^{a+b} {{a+b}\choose{k}} a^{a+b-k} \ b^{k}$

The left-hand side of the inequality equals the term in the sum on the right side with $k=b$, so the result follows.

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- Feb 14, 2012

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It seems to me that solving or proving any given inequalities problems is your strong suit!

Thanks for participating by the way!