$VT=\frac{1}{ab+bc+ca}+\left[ \frac 1{ab+bc+ca} +\frac 1{ab+bc+ca}+\frac 1{a^2+b^2+c^2}\right]$$ \ge \frac{1}{\frac{(a+b+c)^2}3}+\frac 9{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{3}{(a+b+c)^2}+\frac{9}{(a+b+c)^2} =12$
Dấu "=" xảy ra $\Leftrightarrow a=b=c=\frac 13$