Question

cm bdt \(\frac{a+b}{2}-\sqrt{ab}< \frac{\left(a-b\right)^2}{8b}\left(a>b>0.\right)\)

Answer 1

\(\frac{a+b}{2}-\sqrt{ab}=\frac{a-2\sqrt{ab}+b}{2}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}=\frac{4b\left(\sqrt{a}-\sqrt{b}\right)^2}{8b}\)

\(=\frac{\left(2\sqrt{b}\right)^2\left(\sqrt{a}-\sqrt{b}\right)^2}{8b}=\frac{\left(2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\right)^2}{8b}=\frac{\left(2\sqrt{ab}-2b\right)^2}{8b}\)

vì \(0< =\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\Rightarrow2\sqrt{ab}< =a+b\Rightarrow2\sqrt{ab}-2b< =a+b-2b\)

\(\Rightarrow2\sqrt{ab}-2b< =a-b\)

dấu = xảy ra khi và chỉ khi a=b mà a>b(giả thiết)\(\Rightarrow2\sqrt{ab}-2b< a-b\Rightarrow\frac{\left(2\sqrt{ab}-2b\right)^2}{8b}< \frac{\left(a-b\right)^2}{8b}\)

\(\Rightarrow\frac{a+b}{2}-\sqrt{ab}< \frac{\left(a-b\right)^2}{8b}\left(đpcm\right)\)