Từ BĐT \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)
\(\Leftrightarrow ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)
\(\Rightarrow VT\ge1+\dfrac{3}{\dfrac{\left(a+b+c\right)^2}{3}}\ge\dfrac{6}{a+b+c}\)
Cần chứng minh \(1+\dfrac{3}{\dfrac{t^2}{3}}-\dfrac{6}{t}\ge0\left(t=a+b+c\right)\)\(\Leftrightarrow\dfrac{\left(t-3\right)^2}{t^2}\ge0\)