Question

Biết tổng a,b,c bằng 1. Chứng minh rằng:

 \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

Answer 1

Đề phải cho \(a,b,c>0\)nữa nha

Bài làm :

Áp dụng bđt Cauchy :

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

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

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\frac{1}{abc}}=\frac{9\cdot\sqrt[3]{abc}}{\sqrt[3]{abc}}=9\)

Hay \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)( vì \(a+b+c=1\))

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)