<s>[INST] Is there a simple proof of this inequality? I want to prove that
$$ (a-1)^2+(b-1)^2 \ge 2(\sqrt{ ab}-1)^2 $$ for any positive real numbers $a,b$, and that equality holds iff $a=b$.
Edit: My "proof" for this inequality was wrong! It turns out the inequality holds iff $a=b$ or $\sqrt{a}+\sqrt{b} \ge \sqrt{2}$. (See the answers for details).
 [/INST] $$ (a-1)^2+(b-1)^2 - 2(\sqrt{ab}-1)^2 = a^2+b^2-2ab-2(a+b)+4\sqrt{ab} = (\sqrt{a}-\sqrt{b})^2((\sqrt{a}+\sqrt{b})^2-2), $$
which is nonnegative if and only if $\sqrt{a}+\sqrt{b} \geq \sqrt{2}$ or $a=b$.
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