Cho `a>b>0.CM: (a+b)/2 < (a-b)^2/(8b)+ \sqrt{ab}`

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_