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

$$f'(t)=\frac{t^3+3t-2}{t^3(t-1)^2}$$
Set
$$g(t)=t^3+3t-2$$
$g(0)=-2<0$ and $g(1)=2>0$, then there is $t_0\in (0,1)$ such that $g(t_0)=0$. On the other hand $g'(t)=3t^2+3>0$, therefore $g$ is on-to-one and has one root. 
$$$$

Indeed $f(t_0)$ is minimum of $f$ on $(0,1)$. We have
$$t_0^3+3t_0-2=0\implies t_0(t_0^2+1)=2(1-t_0)\implies \frac{1+t_0^2}{t_0^2(1-t_0)}=\frac{2}{t_0^3}.$$ 
$$f(t_0)=\frac{1+t_0^2}{t_0^2(1-t_0)}=\frac{2}{t_0^3}=\frac{2}{2-3t_0}$$
I can compute $t_0$ by numerical method. Anyway, if $t_0\approx 0.5833$ then
$$f(t)\ge\frac{2}{2-3(0.5833)}>8$$
