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).

With the change of variable $a=e^u, b=e^v$, we are stating that
$$ \frac{f(u)+f(v)}{2}\geq f\left(\frac{u+v}{2}\right) \tag{1}$$
for $f(z) = (e^z-1)^2$ and $u,v\in\mathbb{R}$. Since $f(z)$ is a function of class $C^2$, the midpoint convexity stated in $(1)$ is equivalent to convexity. However
$$ f''(z) = 2e^z(2e^z-1) \tag{2}$$
is not always non-negative, and the original inequality does not hold for $a=\frac{1}{2}, b=\frac{1}{3}$, for instance. However it holds if $a,b\geq \frac{1}{2}$, since the RHS of $(2)$ is non-negative for $z\geq -\log 2$.
