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

Since $\displaystyle 0<t<1, \;\frac{1+t^2}{t^2(1-t)}>8\iff 1+t^2>8t^2(1-t)=8t^2-8t^3\iff8t^3-7t^2+1>0$
If $g(t)=8t^3-7t^2+1,\;\; g^{\prime}(t)=24t^2-14t=0\iff t=0 \text{ or }t=\frac{7}{12}$.
Since $g(\frac{7}{12})=\frac{89}{432}$ is the minimum value of $g$ on $(0,1)$, $\;\;g(t)>0$ for $0<t<1$
