<s>[INST] For $0For $0<b \leq a$ prove that $$\frac{1}{8}\left(\frac{(a-b)^2}{a} \right) \leq \frac{a+b}{2}-\sqrt{ab} \leq \frac{1}{8}\left(\frac{(a-b)^2}{b} \right) $$
I was trying compare the following expressions
$$\frac{a+b}{2}-\sqrt{ab} =\frac{1}{2}(\sqrt{a}-\sqrt{b})^2$$
$$\frac{1}{8}\left(\frac{(a-b)^2}{a} \right)=\frac{1}{2a}\left(\frac{a-b}{2}\right)^2\leq\frac{1}{a}(a-b)^2$$
and use the fact that $\sqrt{a}  \leq a$ when  $a>0$ but I don´t get good results.
Any hint of how I should start?
 [/INST] Hint:
$$
\frac{1}{2}(\sqrt a  - \sqrt b )^2  = \frac{b}{{2(\sqrt a  + \sqrt b )^2 }}\left( {\frac{{(a - b)^2 }}{b}} \right)
$$
and
$$
\frac{1}{2}(\sqrt a  - \sqrt b )^2  = \frac{a}{{2(\sqrt a  + \sqrt b )^2 }}\left( {\frac{{(a - b)^2 }}{a}} \right).
$$
Use the fact that $0<b\leq a$ to estimate the right-hand sides from above and below.
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