Đáp án:
$\frac{a+b}{2} <\frac{(a-b)^2}{8b} +\sqrt{ab}$
Giải thích các bước giải:
$\frac{a+b}{2} <\frac{(a-b)^2}{8b} +\sqrt{ab}$
$⇔a+b < \frac{(a-b)^2}{4b}+2\sqrt{ab}$
$⇔a-2\sqrt{ab} +b< \frac{(a-b)^2}{4b}$
$⇔(\sqrt{a}-\sqrt{b})^2.4b< (a-b)^2$
$⇔(\sqrt{a}-\sqrt{b})^2.4b<[ (\sqrt{a}-\sqrt{b}).(\sqrt{a}+\sqrt{b})]^2$
$⇔4b <(\sqrt{a}+\sqrt{b})^2$
$⇔0 <(\sqrt{a}+\sqrt{b})^2-(2\sqrt{b})^2$
$⇔0< (\sqrt{a}+\sqrt{b}-2\sqrt{b}).(\sqrt{a}+\sqrt{b}+2\sqrt{b})$
$⇔0<(\sqrt{a}-\sqrt{b}).(\sqrt{a}+3\sqrt{b})$
vì a >b > 0 nên$ \sqrt{a}-\sqrt{b}>0 ;(\sqrt{a}+3\sqrt{b})>0$
⇒$(\sqrt{a}-\sqrt{b}).(\sqrt{a}+3\sqrt{b}) >0$
⇒$\frac{a+b}{2} <\frac{(a-b)^2}{8b} +\sqrt{ab}$
_đpcm_
