<s>[INST] 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.
 [/INST] You have to prove $\frac{a^2}{b^2}+\frac{b^2}{a^2} \ge 2$.
Consider that $(\sqrt{x}-\sqrt{y})^2 \ge 0$, since it is a square. This is equivalent to $x+y \ge 2\sqrt{xy}$. Put $x=\frac{a^2}{b^2}$ and $y=\frac{b^2}{a^2}$ and you get the result. [This is simply AM-GM inequality.]
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