Ta có:
\(A=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\frac{9}{a+b+c}=9\)
Dấu = xảy ra khi a = b = c
ap dung nếu cần c/m:\(t+\frac{1}{t}\ge2\) mọi t>0 đẳng thức khi t=1
\(\ge9\) khi a=b=c
Áp dung BĐT Cô - si
\(_{\Rightarrow\hept{\begin{cases}a+b+c\ge\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{cases}}}3\sqrt[3]{abc}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc.\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Vậy GTNN của \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=9\)
