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2. Prove for real numbers $a \neq 0, b$ the inequality

$$a^{2}+b^{2}+\frac{2}{a^{2}}+\frac{2 b}{a} \geqslant 2$$

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

I solved it I didn't Skip

Official solution

2. The solution follows from the following inequality:

$$\left(a-\frac{1}{a}\right)^{2}+\left(b+\frac{1}{a}\right)^{2} \geqslant 0$$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.