How to prove $\prod_{i=1}^{n}\left(\dfrac{x_{i}}{\sin{x_{i}}}\right)^{2a_{i}}+\prod_{i=1}^{n}\left(\dfrac{x_{i}}{\tan{x_{i}}}\right)^{a_{i}}>2$? I wish to show that 

$$\prod_{i=1}^{n}\left(\dfrac{x_{i}}{\sin{x_{i}}}\right)^{2a_{i}}+\prod_{i=1}^{n}\left(\dfrac{x_{i}}{\tan{x_{i}}}\right)^{a_{i}}>2,$$

for a positive integer $n$, $x_{i}\in(0,\dfrac{\pi}{2}),$ and $a_{i}\ge 1$ 
I unsuccessfully tried to solve this problem using Bernoulli inequality

$$(1+x)^n\ge 1+nx,n\ge1,x>-1$$

I am thankful to everyone who can solve it. 
This problem is from $2010$ China Maths Olympic team selection exercise, it is from this word problem $11$. 
I think that the problem is nice and maybe not easy. I thank to everyone who can help to solve it.
 A: It seems the following.
The proof of the inequality is simple and follows by induction from the following simple inequalities holding for 
each $x\in\left(0,\frac{\pi}2\right)$:
$\left(\frac x{\sin x}\right)^2\ge \frac x{\sin x}\ge 1\ge \frac x{\tan x}$. 
Proof. These inequalities should be well known or they can be derived from the inequality $\sin x<x<\tan x$, which can be proved by the way $f(x)>0$ for each $x$ provided $f(0)=0$ and $f’(x)>0$ for each $x$, where $f(x)=x-\sin x$ or $f(x)=\tan x-x$.$\square$
$\left(\frac x{\sin x}\right)^2+\frac x{\tan x}>2$.
Proof. By routine equivalent transformations we can reduce it to the inequality $f(x)=x^2+x\sin x\cos x-2\sin^2 x>0$. Now we have $f’(x)=3x-2x\sin^2 x-3\sin x\cos x$, $f’’(x)=4\sin x(\sin x-x\cos x)>0$ and $f(0)=f’(0)=0$.$\square$ 
Proof.  If $u,v>0$, $a>1$  and $u+v>2$ then $u^a+v^a>2$. Put $f(a)= u^a+v^a – 2$. Then $f(1)>0$ and $f’(a)=a(u^{a-1}+v^{a-1})>0$.$\square$
If $u_1\ge v_1>0$,  $u_2\ge v_2>0$, $u_1+v_1>2$ and $u_2+v_2>2$ then $u_1u_2+v_1v_2>2$.$\square$
Proof.  By Rearrangement inequality,  $u_1u_2+v_1v_2\ge u_1v_2+v_1u_2$. Then
$u_1u_2+v_1v_2\ge (u_1u_2+v_1v_2+u_1v_2+v_1u_2)/2=(u_1+v_1)(u_2+v_2)/2>2$.
