BĐT $\Leftrightarrow \frac{1+abc}{a(b+1)}+\frac{1+abc}{b(c+1)}+\frac{1+abc}{c(1+a)}\geq3$
$\Leftrightarrow \left[ {\frac{1+abc}{a(b+1)}+1} \right]+\left[ {\frac{1+abc}{b(1+c)}}+1 \right]+\left[ {\frac{1+abc}{c(1+a)}}+1 \right]\geq 6$
$\Leftrightarrow \frac{(1+a)+ab(1+c)}{a(1+b)}+\frac{(1+b)+bc(1+a)}{b(1+c)}+\frac{(1+c)+ca(1+b)}{c(1+a)}\geq 6$
$\Leftrightarrow \left[ {\frac{1+a}{a(1+b)}+\frac{a(1+b)}{1+a}} \right]+\left[ {\frac{1+b}{b(1+c)}+\frac{b(1+c)}{1+b}} \right]+\left[ {\frac{1+c}{c(1+a)}+\frac{c(1+a)}{1+c}} \right]\geq 6$
dến đây thỳ ngu đến mấy cũng dễ thấy $VT\geq6$ $\Rightarrow$ DPCM