Question

CM: \(\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}\ge\sqrt{a}+\sqrt{b}\)

Answer 1

Bổ xung đk : \(a;b>0\)

Theo bđt Cauchy ta có :

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

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

\(\Rightarrow\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}+\sqrt{a}+\sqrt{b}\ge2\sqrt{a}+2\sqrt{b}\)

\(\Rightarrow\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}\ge\sqrt{a}+\sqrt{b}\)(đpcm)