Question

Cho a,b > 0 phân biệt tùy ý. Đặt \(A=\frac{a+b}{2};B=\sqrt{ab}\). CMR: \(B< \frac{\left(a-b\right)^2}{8\left(A-B\right)}< A\)

Answer 1

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

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

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