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).
 A: 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 $z\geq -\log 2$.
A: $$ (a-1)^2+(b-1)^2 - 2(\sqrt{ab}-1)^2 = a^2+b^2-2ab-2(a+b)+4\sqrt{ab} = (\sqrt{a}-\sqrt{b})^2((\sqrt{a}+\sqrt{b})^2-2), $$
which is nonnegative if and only if $\sqrt{a}+\sqrt{b} \geq \sqrt{2}$ or $a=b$.
A: A sort of "working backwards" solution, but the steps are all reversible: Rewrite as $$a^2-2a+1 + b^2-2b+1 \geq 2ab - 4\sqrt{ab} + 2$$
So $$a^2 -2ab + b^2 \geq 2a + 2b - 4 \sqrt{ab}$$
$$(a-b)^2 \geq 2(a - 2\sqrt{ab} + b)$$
$$(a-b)^2 \geq 2(\sqrt{a}-\sqrt{b})^2$$
In the case where $a = b$, we have equality. In the case where $a \neq b$, we can divide by $\sqrt{a}-\sqrt{b}$:
$$(\sqrt{a} + \sqrt{b})^2 \geq 2$$
$$\sqrt{a} + \sqrt{b} \geq \sqrt{2}$$ This last step follows since the quantity $\sqrt{a} + \sqrt{b}$ is nonnegative.
I got stuck on this part but Jack D'Aurizio's answer above helps; the inequality should hold whenever this is true (each other step is invertible).
