Minimizing $\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2$ While solving a problem I came across this task, minimizing 
\begin{align}
\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2.
\end{align}
One can easily do it with calculus to show that the minimum value is $12.5$.
I tried to do it using trigonometric identities and fundamental inequalities (like AM-GM, Cauchy-Schwarz, etc.) but failed. Can someone help me to do it using trig identities and inequalities?

Another method:
$$\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2=\\
\sin ^4x+\cos ^4x+4+\frac1{\sin^4 x}+\frac1{\cos^4 x}=\\
(\sin ^2x+\cos ^2 x)^2-2\sin^2x\cos^2x+4+\frac{(\sin ^2x+\cos ^2 x)^2-2\sin^2x\cos^2x}{\sin^4x\cos^4x}=\\
1-\frac12\sin^2 2x+4+\frac{1-\frac12\sin 2x}{\frac1{16}\sin^4 2x}=\\
\left(1-\frac12\sin^2 2x\right)\left(1+\frac{16}{\sin^4 2x}\right)+4\ge \\
\left(1-\frac12\cdot 1\right)\left(1+\frac{16}{1^2}\right)+4=12.5.
$$
Note: $1-\frac12\sin ^2 2x>0$, so $\sin^2 2x$ should be maximized to $1$.
