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
 A: $\bf{My\; Solution::}$ 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 $$\frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$$
A: 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 $0<t<1$
A: 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
A: $$f'(t)=\frac{t^3+3t-2}{t^3(t-1)^2}$$
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
$$f(t)\ge\frac{2}{2-3(0.5833)}>8$$
A: Your original derivation was correct. Now, you have to take the derivative of $\frac{1+t^2}{t^2(1-t)}$ and find when it is equal to zero. Then, take the second derivative at that point; if the second derivative is negative, then you have a minimum. Find the value of the function at that point, and if it's larger than 8, then you've proven it.
A: In the same spirit as Behrouz Maleki's answer, if you apply Cardano method to $$t^3+3t-2=0$$ you should notice that there is only one real root given by $$t_*=\sqrt[3]{1+\sqrt{2}}-\frac{1}{\sqrt[3]{1+\sqrt{2}}}\approx 0.596072 $$ which does not seem to be easily simplified.
At this point, the evaluation of $h(t_*)$ and $$h''(t_*)=\frac{2 (t_*+3)}{t_*^4}-\frac{4}{(t_*-1)^3}$$ need to be performed numerically to get $h(t_*)\approx 9.44354$ and $h''(t_*)\approx 117.667$. Using $t_*=0.6$ is more than likely sufficient.
