\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\)\(a+b+c\ge3\left(\frac{ab+bc+ca}{a+b+c}\right)\)
\(\Leftrightarrow\)\(a+b+c\ge3\left(\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}\right)\)
\(\Leftrightarrow\)\(a+b+c\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)