Minimize $x^4+ \frac{1}{x^4} +y^4+ \frac{1}{y^4}$

Question

Minimize $${x^4+ \frac{1}{x^4} +y^4+ \frac{1}{y^4}} $$ Subject to $$x^2+y^2=4$$

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I've managed to solve this by substituting $x=(2+t)^\frac{1}{2}$ and $y=(2-t)^\frac{1}{2}$ and I arrived at a minimum of $\frac{17}{2}$ when $t=0$, but it's kind of tedious.
So I'm asking if there are straightforward ways of finding the minimum using inequalities and not using symmetry.

Answer 1

Let $f(x)=x^2+\frac{1}{x^2}$.

Thus, $f''(x)>0$.

Thus, by Jensen $$x^4+\frac{1}{x^4}+y^4+\frac{1}{y^4}\geq2\left(\left(\frac{x^2+y^2}{2}\right)^2+\frac{1}{\left(\frac{x^2+y^2}{2}\right)^2}\right)=\frac{17}{2}.$$ The equality occurs for $x=y=\sqrt2$, which says that we got a minimal value.

Done!