If $x\in \left(0,\frac{\pi}{4}\right)$ then $\frac{\cos x}{(\sin^2 x)(\cos x-\sin x)}>8$ 
If $\displaystyle x\in \left(0,\frac{\pi}{4}\right)\;,$ Then prove that $\displaystyle \frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$

$\bf{My\; Try::}$ Let $$f(x) = \frac{\cos x}{\sin^2 x(\cos x-\sin x)}=\frac{\sec^2 x}{\tan^2 x(1-\tan x)} = \frac{1+\tan^2 x}{\tan^2(1-\tan x)}$$
Now Put $\tan x= t \in (0,1)\;,$ Then $$h(t) = \frac{1+t^2}{t^2(1-t)}\;\;, 0<t<1 $$  where $h(t)=f(x)$.
Now How can i solve it after that, Help Required ,Thanks

We are to prove that $$\frac{\cos x}{\sin x (\cos x - \sin x)}> 8 \sin x $$
By Cauchy-Schwarz Inequality, since all quantities involved are positive
$$\bf{LHS = }\frac{1}{\cos x} + \frac{1}{\cos x-\sin x} \ge \frac{4}{\sin x + \cos x - \sin x} = \frac{4}{\sin x}$$ 
For the given range of x, we have $1> 2 \sin^2 x$
So $$\frac{4}{\sin x} > \frac{4}{\sin x} \times 2 \sin^2 x = 8 \sin x$$
and we are done
