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.
 A: Apply AM-GM inequalities twice: Let $x_i = \cos \alpha_i \to 0 \leq x_i \leq 1 \to P = x_1x_2\cdots x_n \to P^2 = (1-x_1^2)(1-x_2^2)\cdots (1-x_n^2)\leq \left(\dfrac{n-(x_1^2+x_2^2+\cdots x_n^2)}{n}\right)^n\leq \left(\dfrac{n-n\sqrt[n]{x_1^2x_2^2\cdots x_n^2}}{n}\right)^n= \left(1-\sqrt[n]{P^2}\right)^n\to \sqrt[n]{P^2} \leq 1-\sqrt[n]{P^2}\to \sqrt[n]{P^2} \leq \dfrac{1}{2}\to P^2 \leq 2^{-n}\to P \leq 2^{-\frac{n}{2}}$
A: $\cot\alpha_1.\cot\alpha_2......\cot\alpha_n=1$
$\cos\alpha_1.\cos\alpha_2......\cos\alpha_n=\sin\alpha_1.\sin\alpha_2......\sin\alpha_n$....................(1)
Let $y=\cos\alpha_1.\cos\alpha_2......\cos\alpha_n$(to be maximized)
$\Rightarrow y^2=\cos^2\alpha_1.\cos^2\alpha_2......\cos^2\alpha_n$
$y^2=(\cos\alpha_1.\cos\alpha_2......\cos\alpha_n)(\sin\alpha_1.\sin\alpha_2......\sin\alpha_n)$
$y^2=\frac{1}{2^n}[\sin(2\alpha_1).\sin(2\alpha_2)......\sin(2\alpha_n)]$
As $0\leq\alpha_1,\alpha_2,.....,\alpha_n\leq\frac{\pi}{2}$
$0\leq2\alpha_1,2\alpha_2,.....,2\alpha_n\leq\pi$
$\Rightarrow 0\leq \sin(2\alpha_1),\sin(2\alpha_2),......,\sin(2\alpha_n)\leq 1$
$y^2\leq\frac{1}{2^n}\times 1\Rightarrow y\leq \frac{1}{2^{n/2}}$
Therefore maximum value of $y$ is $\frac{1}{2^{n/2}}$.
A: 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 $$\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdots \cos \alpha_{n}\leq \frac{1}{2^{\frac{n}{2}}}$$
