Question

Cho a > 0, b > 0. CMR: \(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)

Answer 1

Áp dụng BĐT AM-GM ta có:

\(\frac{a}{\sqrt{b}}+\sqrt{b}\ge2.\sqrt{\frac{a}{\sqrt{b}}.\sqrt{b}}=2\sqrt{a}\)

Tương tự:\(\frac{b}{\sqrt{a}}+\sqrt{a}\ge2\sqrt{\frac{b}{\sqrt{a}}.\sqrt{a}}=2\sqrt{b}\)

Cộng theo vế BĐT ta được:\(\frac{a}{\sqrt{b}}+\sqrt{b}+\frac{b}{\sqrt{a}}+\sqrt{a}\ge2\left(\sqrt{a}+\sqrt{b}\right)\)

\(\Rightarrow\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\)