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.

By Minkowski (see here: https://en.wikipedia.org/wiki/Minkowski_inequality) we obtain
$$\sqrt[4]{ \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta}}+\sqrt[4]{ \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta}}\geq\sqrt[4]{\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4}.$$
Thus, it remains to prove that
$$\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4\geq68.$$
Now, let $\sin\theta+\cos\theta=2k\sqrt{\sin\theta\cos\theta}$.
Thus, $k\geq1$ and we need to prove that
$$\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2(\sin^2\theta+\cos^2\theta)^2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4\geq68(\sin^2\theta+\cos^2\theta)$$ or
$$(k+1)^2+16(2k^2-1)^2(k+1)^2\geq68(2k^2-1)$$ or
$$(k-1)(64k^5+192k^4+192k^3+64k^2-119k-85)\geq0,$$
which is obvious.
Done!
