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?)

  • $\begingroup$ What I'd try: Put $\cos^2\alpha=:p$, $\sin^2\alpha=:q$. Then $p$, $q$ are both nonnegative, and $p+q=1$. $\endgroup$ – Christian Blatter Aug 15 '14 at 15:42
  • $\begingroup$ @ChristianBlatter,i wrote the two am-gm inequalities and used that fact.did i missed your point? $\endgroup$ – user2838619 Aug 15 '14 at 15:49

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

$$\frac{m\cdot\dfrac{\sin^2\alpha}m+n\cdot\dfrac{\cos^2\alpha}n}{m+n}\ge \left[\left(\dfrac{\sin^2\alpha}m\right)^m\left(\dfrac{\cos^2\alpha}n\right)^n\right]^{\dfrac1{m+n}}$$

  • $\begingroup$ @user2838619, Thanks your observation $\endgroup$ – lab bhattacharjee Aug 15 '14 at 16:33

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