Đặt \(A=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=\left(a+b+c\right).\frac{1}{a}+\left(a+b+c\right).\frac{1}{b}+\left(a+b+c\right).\frac{1}{c}\)
\(=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)
\(=\frac{a}{a}+\frac{b+c}{a}+\frac{b}{b}+\frac{a+c}{b}+\frac{c}{c}+\frac{a+b}{c}\)
\(=1+\frac{b+c}{a}+1+\frac{a+c}{b}+1+\frac{a+b}{c}\)
\(=3+\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)
Ta có: trong 1 tam giác thì tổng độ dài 2 cạnh bao giờ cũng lớn hơn cạnh còn lại ( bất đẳng thức tam giác )
\(\Rightarrow\hept{\begin{cases}b+c>a\\a+c>b\\a+b>c\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{b+c}{a}>1\\\frac{a+c}{b}>1\\\frac{a+b}{c}>1\end{cases}}\)
\(\Rightarrow A>3+1+1+1\)
\(\Rightarrow A>6\left(đpcm\right)\)
