Maximum of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}....\cos \alpha_{n}.$ 
Maximum value of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}.$ If it is given that


$\cot \alpha_{1}\cdot \cot \alpha_{2}\cdot \cot \alpha_{3}.......\cot \alpha_{n}  = 1$ and $\displaystyle 0 \leq \alpha_{1},\alpha_{2},\alpha_{3},.......,\alpha_{n}\leq \frac{\pi}{2}.$

$\bf{My\; Try::}$ Using $\bf{A.M\geq G.M}$
$$\displaystyle \frac{\sin^2 \alpha_{1}+\sin^2 \alpha_{2}+\sin^2 \alpha_{3}+........+\sin^2 \alpha_{n}}{n}\geq \sqrt[n]{\sin^2 \alpha_{1}\cdot \sin^2 \alpha_{2}...\sin^2 \alpha_{n}}$$
So we get
$$\displaystyle \sin^2 \alpha_{1}+\sin^2 \alpha_{2}+\sin^2 \alpha_{3}+........+\sin^2 \alpha_{n}\geq n\cdot \sqrt[n]{\sin^2 \alpha_{1}\cdot \sin^2 \alpha_{2}...\sin^2 \alpha_{n}}.......(1)$$
Similarly
$$\displaystyle \frac{\cos^2 \alpha_{1}+\cos^2 \alpha_{2}+\cos^2 \alpha_{3}+........+\cos^2 \alpha_{n}}{n}\geq \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}$$
So we get
$$\displaystyle \cos^2 \alpha_{1}+\cos^2 \alpha_{2}+\cos^2 \alpha_{3}+........+\cos^2 \alpha_{n}\geq n\cdot \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}.......(2)$$
Now Adding $(1)$ and $(2)\;,$ We get $$n\geq 2n\cdot \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}$$
Because above it is given that $$\sin \alpha_{1}\cdot \sin\alpha_{2}\cdot\sin \alpha_{3}.........\sin\alpha_{n}=\cos\alpha_{1}
\cdot\cos\alpha_{2}\cdot \sin\alpha_{3}......\cos\alpha_{n}$$
So we get $$\displaystyle \cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}\leq \frac{1}{2^{\frac{n}{2}}}$$
My question is that can we solve it using any other short method? If yes then please explain it here.
Thanks.

Using $$\sec^2 \alpha_{1}\cdot \sec^2 \alpha_{2}\cdot \sec^2 \alpha_{3}\cdots \sec^2 \alpha_{n} =(1+\tan^2 \alpha_{1})\cdot (1+\tan^2 \alpha_{2})\cdots \cdots (1+\tan^2 \alpha_{n})$$
Now Apply $\bf{A.M\geq G.M}$
$$(1+\tan^2 \alpha_{1})\cdot (1+\tan^2 \alpha_{2})\cdots \cdots (1+\tan^2 \alpha_{n})\geq 2\tan \alpha_{1}\cdot 2\tan \alpha _{2}\cdot \cdots2\tan \alpha_{n}=2^{n}$$
So $$\sec^2 \alpha_{1}\cdot \sec^2 \alpha_{2}\cdot \sec^2 \alpha_{3}\cdots \sec^2 \alpha_{n}\geq 2^n$$
So $$\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdots \cos \alpha_{n}\leq \frac{1}{2^{\frac{n}{2}}}$$
