AM-GM Inequality 1

$f(x)=\frac{1}{\cos^2{x}}+\frac{4}{\sin^2{x}}$

Due to the Pythagorean Trigonometric Identity,

$f(x)=\frac{\sin^2{x}+\cos^2{x}}{\cos^2{x}}+\frac{4(\sin^2{x}+\cos^2{x})}{\sin^2{x}}=1+\frac{\sin^2{x}}{\cos^2{x}}+4+\frac{4\cos^2{x}}{\sin^2{x}}$

Due to the Arithmetic Mean-Geometric Mean Inequality,

$\frac{\sin^2{x}}{\cos^2{x}}+\frac{4\cos^2{x}}{\sin^2{x}}\geq 2\sqrt{\frac{\sin^2{x}}{\cos^2{x}}\times\frac{4\cos^2{x}}{\sin^2{x}}}=4$

$\therefore f(x)\geq 9$

A direct application of Titu's Lemma gives us

$\frac{1}{\cos^2x}+\frac{4}{\sin^2x}\ge\frac{\left(1+2\right)^2}{\cos^2x+\sin^2x}=9$

Equality holds when

$\frac{1}{\cos x}=\frac{2}{\sin x}$

i.e. for $\displaystyle x=\arctan\left(2\right)$