Feb 10, 2014 Thread starter Admin #1 anemone MHB POTW Director Staff member Feb 14, 2012 3,965 Let $a$ and $b$ be positive integers. Show that $\dfrac{(a+b)!}{(a+b)^{a+b}}\le \dfrac{a! \cdot b!}{a^ab^b}$.
Let $a$ and $b$ be positive integers. Show that $\dfrac{(a+b)!}{(a+b)^{a+b}}\le \dfrac{a! \cdot b!}{a^ab^b}$.
Feb 11, 2014 #2 P Petek Active member Jan 8, 2013 25 Spoiler 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.
Spoiler 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.
Feb 13, 2014 Thread starter Admin #3 anemone MHB POTW Director Staff member Feb 14, 2012 3,965 Hi Petek, It seems to me that solving or proving any given inequalities problems is your strong suit! Thanks for participating by the way!
Hi Petek, It seems to me that solving or proving any given inequalities problems is your strong suit! Thanks for participating by the way!