$VT=\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{ba+bc}+\frac{\frac{1}{c^2}}{ca+cb}\geq \frac{\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2}{2(ab+bc+ca)}=\frac{(ab+bc+ca)^2}{2(ab+bc+ca)}\geqslant \frac{ab+bc+ca}{2} \geq \frac{3\sqrt[3]{(abc)^2}}{2}=\frac{3}{2}$dấu$"="\Leftrightarrow a=b=c=1$
$VT=\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{ba+bc}+\frac{\frac{1}{c^2}}{ca+cb}\geq \frac{\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2}{2(ab+bc+ca)}=\frac{(a+b+c)^2}{2(ab+bc+ca)}\geqslant \frac{3(ab+bc+ca)}{2(ab+bc+ca)}=\frac{3}{2}$dấu$"="\Leftrightarrow a=b=c=1$
$VT=\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{ba+bc}+\frac{\frac{1}{c^2}}{ca+cb}\geq \frac{\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2}{2(ab+bc+ca)}=\frac{(a
b+b
c+c
a)^2}{2(ab+bc+ca)}\geqslant \frac{ab+bc+ca}{2
} \geq \frac{3\sqrt[3]{(abc)
^2}}{2}=\frac{3}{2}$dấu$"="\Leftrightarrow a=b=c=1$