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Inequality Challenge V

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,681
Let $a$ and $b$ be positive integers. Show that $\dfrac{(a+b)!}{(a+b)^{a+b}}\le \dfrac{a! \cdot b!}{a^ab^b}$.
 

Petek

Member
Jan 8, 2013
20
Rewrite the inequality as

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

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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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,681
Hi Petek,

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

Thanks for participating by the way!
 

Petek

Member
Jan 8, 2013
20
Thanks for posing such interesting problems!