Ta có $VT= \frac{a^{2}b^{2}}{abc}+\frac{b^{2}c^{2}}{abc}+\frac{c^{2}a^{2}}{abc}=\frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}{abc}\geqslant \frac{abc(a+b+c)}{abc}=a+b+c$ BĐT phụ : $x^{2}+y^{2}+z^{2}\geqslant xy+yz+zx $ (CM Đơn giản)
$\Rightarrow $ $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} \geqslant ab^{2}c+bc^{2}a+ca^{2}b = abc(a+b+c)$ $(dpcm)$