Áp dụng BĐT: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(VT=\left(\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\right)+2\left(\dfrac{1}{b+c-a}+\dfrac{1}{a+b-c}\right)+3\left(\dfrac{1}{a+c-b}+\dfrac{1}{a+b-c}\right)\\ \Leftrightarrow VT\ge\dfrac{4}{2c}+2\cdot\dfrac{4}{2b}+3\cdot\dfrac{4}{2a}=\dfrac{6}{a}+\dfrac{4}{b}+\dfrac{2}{c}\)
Dấu \("="\Leftrightarrow a=b=c\)

