Q. The maximum value of $\cos x_1\cdot \cos x_2 \cdot \cos x_3...\cos x_n$ under restrictions $0\le x_1,x_2,x_3,...,x_n \le \frac{\pi}{2}$ and $\cot x_1 \cdot \cot x_2 \cdot \cot x_3...\cot x_n=1$.
MY ATTEMPT:
From question, we get $$\cos x_1 \cdot \cos x_2 \cdot \cos x_3...\cos x_n=\sin x_1 \cdot \sin x_2 \cdot \sin x_3...\sin x_n ...(1)$$ Using AM,GM inequality, we get: $$\frac{\cos x_1+\cos x_2+\cos x_3+...+\cos x_n}{n}=\sqrt[n]{\cos x_1\cdot \cos x_2 \cdot \cos x_3...\cos x_n}\,\,\,...(2)$$ $$\frac{\sin x_1 +\sin x_2+\sin x_3+...+\sin x_n}{n}=\sqrt[n]{\sin x_1 \cdot \sin x_2 \cdot \sin x_3...\sin x_n}\,\,\,...(3)$$
The RHS term in equations $(2)\,\,,(3)$ are equal by using equation $(1)$. So we get: $$\cos x_1+\cos x_2+\cos x_3+...+\cos x_n=\sin x_1 +\sin x_2+\sin x_3+...+\sin x_n$$ Now i notice that if i somehow prove that $\sin x_1=\cos x_1,\,\,\sin x_2=\cos x_2,...,\,\,\sin x_n=\cos x_n$, so i get every angle as '$\frac\pi 4$' $$\cos x_1 \cdot \cos x_2 \cdot \cos x_3...\cos x_n=\biggl(\frac{1}{\sqrt 2}\biggr)^n=\biggl(\frac{1}{ 2^\frac{n}{2}}\biggr)$$; Which is the answer. But i am unsure how to prove such thing or is it even possible with how much i have attempted. Also if possible, give any other solution to the question which is shorter/interesting.
Using Arithmetic Geomeyric Inequality
$\displaystyle \sec^2(x_{i})=1+\tan^2(x_{i})\geq 2\tan(x_{i})$
$\displaystyle \prod^{n}_{i=1}\sec^2(x_{i})\geq \prod^{n}_{i=1}2\tan(x_{i})=2^{n}$
$\displaystyle \prod^{n}_{i=1}\cos(x_{i})\leq \frac{1}{2^{\frac{n}{2}}}$