Proving $ \frac{(a-b)^2}{8a}\leqslant\frac{a+b}{2} - \sqrt{ab} \leqslant\frac{(a-b)^2}{8b}$ when $ 0<b\le a$

Question

I'd like to prove that $$\frac{(a-b)^2}{8a}\leqslant\frac{a+b}{2} - \sqrt{ab}\leqslant\frac{(a-b)^2}{8b}$$ when $ 0<b\leqslant a\,$.

I have been brushing up on inequalities and I cannot figure this one out.
I have searched this site, and found this Show that $\frac{a+b}{2} \ge \sqrt{ab}$ for $0 \lt a \le b$ , but haven't seen how to apply it.
I have tried to find a factor that I can multiply the entire inequality with that will allow me to simplify the middle term. So far no luck there.

I have also tried assuming that $b > a$ and looked for a contradiction to either of the terms individually.

Does anyone have any hints that can get me somewhere?

Answer 1

We have $$ \frac{a+b}{2}-\sqrt{ab}=\frac{(\sqrt{b}-\sqrt{a})^2}{2} =\frac{(b-a)^2}{2(\sqrt{a}+\sqrt{b})^2}. $$ Then you can get your result using $2\sqrt{b}\leq \sqrt{a}+\sqrt{b} \leq 2\sqrt{a}$.

Answer 2

For the first inequality: $$\frac{(a-b)^2}{8a} \leq \frac{a+b}{2} - \sqrt{ab} = \frac{(\sqrt a - \sqrt b)^2} {2}$$

$$ \Leftrightarrow 0 \leq \frac{a-b}{2\sqrt a} \leq \sqrt a - \sqrt b \Leftrightarrow \frac{\sqrt a + \sqrt b}{2\sqrt a} \leq 1$$

Similar for the second inequality.