Question

Answer 1

a: \(a+b+c\ge3\cdot\sqrt[3]{abc}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\cdot\sqrt[3]{\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}}=3\cdot\sqrt[3]{\frac{1}{abc}}=3\cdot\frac{1}{\sqrt[3]{abc}}\)

Do đó: \(P=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\cdot\sqrt[3]{abc}\cdot3\cdot\frac{1}{\sqrt[3]{abc}}=9\forall a,b,c>0\)

Dấu '=' xảy ra khi a=b=c