\(VT\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}=\frac{\left(a+b+c\right)^2}{abc\left(a+b+c\right)}\ge\frac{3\left(a+b+c\right)^2}{\left(ab+bc+ca\right)^2}\) (1)
Mặt khác:
\(\left(a+b+c\right)^2=\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)\ge3\sqrt[3]{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2}\)
\(\Leftrightarrow\left(a+b+c\right)^6\ge27\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^6}{27\left(ab+bc+ca\right)^2}\ge a^2+b^2+c^2\Leftrightarrow\frac{\left(a+b+c\right)^2.3^4}{27\left(ab+bc+ca\right)^2}\ge a^2+b^2+c^2\)
\(\Leftrightarrow\frac{3\left(a+b+c\right)^2}{\left(ab+bc+ca\right)^2}\ge a^2+b^2+c^2\) (2)
(1);(2) \(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge a^2+b^2+c^2\)
