Prove $\left|\frac{2a}{b} + \frac{2b}{a} \right| \ge 4$ for all nonzero $a,b$ 
Prove $\left|\frac{2a}{b} + \frac{2b}{a} \right| \ge 4$ for all nonzero $a,b$

I started by squaring both sides and proving:
$(\mid\frac{2a}{b} + \frac{2b}{a} \mid)^2 \ge 4^2, \forall a,b \in \mathbb{R} - \left\{0\right\}$
My work:
Consider: $(\mid\frac{2a}{b} + \frac{2b}{a} \mid)^2$
$ = (\frac{2a}{b} + \frac{2b}{a})^2 $
$ = \frac{4a^2}{b^2} + 8 + \frac{4b^2}{a^2}$
$ = \frac{4(a^4 + b^4)}{a^2b^2} + 8$
This is where I get stuck. How do I prove:
$\frac{4(a^4 + b^4)}{a^2b^2} + 8 \ge 16$
Or I could simplify it to:
$\frac{a^2}{b^2} + \frac{b^2}{a^2} \ge 2$
But I am still not sure how to prove the above.

Note that this is just
$$\left|\dfrac{2}{x} + 2x\right|\ge 4$$
Now squaring both sides like you suggest gives
$$\dfrac{4}{x^2}+8+4x^2\ge 16$$
$$\iff \dfrac{4}{x^2}-8+4x^2\ge 0$$
$$\iff \left(\dfrac{2}{x} - 2x\right)^2\ge 0$$
