TITLE: Prove that : $f(\sin x)+f(\cos x) \ge 196, \forall x\in\left(0;\frac{\pi}{2}\right)$
QUESTION [6 upvotes]: Given: $$f(\tan2x)=\tan^{4}x+\frac{1}{\tan^{4}x}, \forall x\in\left(0;\frac{\pi}{4}\right)$$
Prove that :$f(\sin x)+f(\cos x) \ge 196, \forall x\in\left(0;\frac{\pi}{2}\right)$
Could someone help me ?

REPLY [5 votes]: since
$$t=\tan{2x}=\dfrac{2\tan{x}}{1-\tan^2{x}}=-\dfrac{2}{\tan{x}-\dfrac{1}{\tan{x}}},t>0$$
so
$$\tan{x}-\dfrac{1}{\tan{x}}=-\dfrac{2}{t}$$
and
$$\tan^4{x}+\dfrac{1}{\tan^4{x}}=\left[\left(\tan{x}-\dfrac{1}{\tan{x}}\right)^2+2\right]^2-2$$
so
$$f(t)=\left[\dfrac{4}{t^2}+2\right]^2-2=\dfrac{16}{t^4}+\dfrac{16}{t^2}+2$$
so
$$f(\sin{x})+f(\cos{x})=16\left(\dfrac{1}{\sin^4{x}}+\dfrac{1}{\cos^4{x}}\right)+16\left(\dfrac{1}{\sin^2{x}}+\dfrac{1}{\cos^2{x}}\right)+4$$
since use Holder inequality we have
$$\left(\dfrac{1}{\sin^4{x}}+\dfrac{1}{\cos^4{x}}\right)(\sin^2{x}+\cos^2{x})^2\ge (1+1)^3$$
$$\Longrightarrow \dfrac{1}{\sin^4{x}}+\dfrac{1}{\cos^4{x}}\ge 8$$
$$\left(\dfrac{1}{\sin^2{x}}+\dfrac{1}{\cos^2{x}}\right)(\sin^2{x}+\cos^2{x})\ge (1+1)^2$$
$$\Longrightarrow  \dfrac{1}{\sin^2{x}}+\dfrac{1}{\cos^2{x}}\ge 4$$
so 
$$f(\sin{x})+f(\cos{x})\ge 196$$