Find the smallest value of $f(x) := \left({1\over9}+{32\over \sin(x)}\right)\left({1\over32}+{9\over \cos(x)}\right)$ on the interval $(0,\pi/2)$ 
There's a function defined as:
  $$f(x) := \left({1\over9}+{32\over \sin(x)}\right)\left({1\over32}+{9\over \cos(x)}\right)$$
  In interval $$\left(0,\frac{\pi}{2}\right)$$
  Find the smallest value (Save only its integer value)

I've managed to come to this
$${1\over288}+{{2(\sin(x)+\cos(x))+576}\over(\sin(x)+\cos(x))^2-1}$$
How can I find the smallest value now?
 A: Since both $\cos$ and $\sin $ are positive in $(o,{\pi\over 2})$ we can use Cauchy inequaliy: 
$$ (a^2+b^2)(c^2+d^2)\geq (ac+bd)^2$$
$$\bigg({1\over9}+{32\over \sin(x)}\bigg)\bigg({1\over32}+{9\over \cos(x)}\bigg)\geq \bigg({1\over \sqrt{288}}+{\sqrt{288}\over \sqrt{\sin(x)\cos(x)}}\bigg)^2\geq \bigg({1\over 12\sqrt{2}}+24\bigg)^2$$
We used here $$\sin(x)\cos(x)= {1\over 2}\sin (2x) \leq  {1\over 2}$$
with equality at $x={\pi \over 4}$. So $$y_{min} = \bigg({1\over 12\sqrt{2}}+24\bigg)^2$$
A: $$1+\frac{288}{\sin x}$$ is a decreasing function in $\left(0,\dfrac\pi2\right)$ and its symmetric 
$$1+\frac{288}{\cos x}$$ is increasing.
Hence the minimum occurs ar $x=\dfrac\pi4$.
A: Here's how to continue with your idea:
Let $\sin{x}+\cos{x}=t$.
Then, by the Cauchy-Schwartz inequality we have $$1<t=\sin{x}+\cos{x}\leq\sqrt{(1^2+1^2)(\sin^2x+\cos^2x)}=\sqrt2,$$
where the equality occurs for $x=\frac{\pi}{4},$  and since $$\left(\frac{t+288}{t^2-1}\right)'=-\frac{t^2+576t+1}{(t^2-1)^2}<0,$$ we obtain
$$
f(x)
=\frac{1}{288}+\frac{2(t+288)}{t^2-1}
\ge \frac{1}{288}+\frac{2(\sqrt{2}+288)}{2-1}
=576 + \frac{1}{288} + 2\sqrt{2}$$
and we are done!
