Inequality Challenge V

anemone

MHB POTW Director
Staff member
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

Active member
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.

anemone

MHB POTW Director
Staff member
Hi Petek,

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

Thanks for participating by the way!

Petek

Active member
Thanks for posing such interesting problems!