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

$\bf{My\; Solution::}$ Using $\bf{A.M\geq G.M}$

$$2\sqrt{\sin x(\cos x-\sin x)}\leq \cos x$$

so $$4\sin x(\cos x-\sin x)\leq \cos^2 x\Rightarrow \frac{\sin^2 x(\cos x-\sin x)}{\cos x}\leq \frac{\cos x\sin x}{4}$$

So $$\frac{\sin^2 x(\cos x-\sin x)}{\cos x}\leq \frac{\sin 2x}{8}< \frac{1}{8}$$

So $$\frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$$

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