TITLE: Prove $\sin^{2m}\alpha\cdot\cos^{2n}\alpha\leq\frac{m^m n^n}{(m+n)^{(m+n)}}$
QUESTION [5 upvotes]: If $n$ and $m$ are natural numbers, Prove: $$\sin^{2m}\alpha\cos^{2n}\alpha\leq\frac{m^mn^n}{(m+n)^{(m+n)}}$$
  Additional info:We should only use AM-GM inequality.We can use Trigonometry identities.

Things I have done so far: for reaching something useful about $\sin^{2m}\alpha\cos^{2n}\alpha$, I tried to do$$\sin^2\alpha+\cos^2\alpha\geq 2\sin\alpha\cos\alpha $$
powering to $m$ $$\frac{1}{4}=\left(\frac{\sin^2\alpha+\cos^2\alpha}{2}\right)^m\geq 2\sin^{2m}\alpha\cos^{2m}\alpha$$
and I stuck here.And for $ \dfrac {m^mn^n} {(m+n)^{(m+n)}} $ I don't know what to do.(Binomial theorem maybe?)

REPLY [4 votes]: Consider $m$ number of $\dfrac{\sin^2\alpha}m$ and $n$ number of $\dfrac{\cos^2\alpha}n$
As each of the term $\ge0$ for real $\alpha;$
using  AM, GM inequality 
$$\frac{m\cdot\dfrac{\sin^2\alpha}m+n\cdot\dfrac{\cos^2\alpha}n}{m+n}\ge \left[\left(\dfrac{\sin^2\alpha}m\right)^m\left(\dfrac{\cos^2\alpha}n\right)^n\right]^{\dfrac1{m+n}}$$