<s>[INST] Prove $ ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} )^{1/4} + ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} )^{1/4}\ge 68^{1/4}$ 
Let $0<\theta<\pi/2$. Prove that $$\left ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} \right )^{1/4}+\left ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} \right )^{1/4}\geqslant (68)^{1/4}$$
and find when the equality case holds.

This is a competition math problem. The material used should only cover up to pre-calculus.
So I quickly found out that equality holds when both of the $\sin^2(\theta)$ and $\cos^2(\theta)$ equals to 1/2, but I am not sure how to prove that this equality is true. I tried to substitute for variables and also use trig identities but just can't find out a way to do this. Thank you guys for helping me.
 [/INST] Clearly, we only need to prove the case when $\theta \in (0, \pi/4]$.
Let
\begin{align*}
 A &= \frac{4}{17}\left(\frac{\sin^2 \theta}{2} + \frac{2}{\cos^2\theta}\right), \\
 B &= \frac{4}{17}\left(\frac{\cos^2 \theta}{2} + \frac{2}{\sin^2\theta}\right).
\end{align*}
It suffices to prove that
$$A^{1/4} + B^{1/4} \ge 2$$
or
$$B^{1/4} - 1 \ge 1 - A^{1/4}$$
or
$$\frac{B - 1}{(B^{1/4} + 1)(B^{1/2} + 1)} \ge \frac{1 - A}{(1 + A^{1/4})(1 + A^{1/2})}$$
or
$$\frac{B - 1}{B^{3/4} + B^{1/2} + B^{1/4} + 1}
\ge \frac{1 - A}{1 + A^{1/4} + A^{1/2} + A^{3/4}}.$$
It is easy to prove that
$B \ge 1 \ge A$.
It suffices to prove that
$$\frac{B - 1}{B + B + B + 1}
\ge \frac{1 - A}{1 + A + A + A}$$
or
$$3AB - A - B - 1 \ge 0$$
or (clearing the denominators)
$$(3\sin^4\theta - 3\sin^2\theta + 56)(1 - 2\sin^2\theta)^2 \ge 0$$
which is true, with equality if $\sin^2\theta = 1/2$.
We are done.
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