Question

cho a+b+c=1. Tìm min \(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\)

Answer 1

Theo AM-GM có :

\(9a+\frac{1}{a}\ge2\sqrt{9a.\frac{1}{a}}=6\)

\(9b+\frac{1}{b}\ge2\sqrt{9b.\frac{1}{b}}=6\)

\(9c+\frac{1}{c}\ge2\sqrt{9c.\frac{1}{c}}=6\)

Cộng vế theo vế :

\(9\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge18\)

\(8\left(a+b+c\right)+\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)\ge18\)

\(\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)\ge10\)

Vậy \(Min=10\)

\(\Leftrightarrow a=b=c=\frac{1}{3}\)