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?
Best Answer
Knowing the answer it's not that difficult to get it another way. Let $y = \cos 2x$. We have $\sin^2 x = \frac{1-y}{2}$, $\cos^2 x =\frac{1+y}{2}$. Then \begin{align} \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 + \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 &= \left(\frac{1-y}{2} + \frac{2}{1-y}\right)^2 + \left(\frac{1+y}{2} + \frac{2}{1+y}\right)^2 = \\ &= \frac{y^6+7y^4-y^2+25}{2(1-y^2)^2} = \\ &= \frac{25}{2} + \frac{y^2(y^4-18y^2+49)}{2(1-y^2)^2} = \\ &= \frac{25}{2} + y^2\Big(\frac{1}{2} + \frac{8}{1-y^2} + \frac{16}{(1-y^2)^2}\Big) \end{align} Since $y^2 \le 1$, the expression in the brackets is strictly positive.