Thiếu ĐK : a;b;c > 0
Áp dụng bđt Cauchy - Schwarz ta có :
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9\) có GTNN là 9
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)