$$Theo BĐT Shwarz ta có: \frac{1}{a^{2}\times (1+a)}+\frac{1}{b^{2}\times (1+b)}+\frac{1}{c^{2}\times(1+c) } \geqslant \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{1+a+1+b+1+c}=\frac{(bc+ac+ab)^{2}}{4\times(abc) }\geqslant\frac{\frac{(a+b+c)^{2}}{3}}{4abc}=\frac{3}{4abc}$$
Theo BĐT Shwarz ta có:
$\frac{1}{a^{2}\times (1+a)}+\frac{1}{b^{2}
\times (1+b)}+\frac{1}{c^{2}\times(1+c)}
$ $\geqslant \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{1+a+1+b+1+c}
$=
$\frac{(bc+ac+ab)^{2}}{4\times(abc) }\geqslant\frac{\frac{(a+b+c)^{2}}{3}}{4abc}=\frac{3}{4abc}$