Quy đồng full :)
\(\frac{1}{a\left(1+b\right)}+\frac{1}{b\left(1+c\right)}+\frac{1}{c\left(1+a\right)}\ge\frac{3}{1+abc}\)
\(\Leftrightarrow\frac{1+abc}{a\left(1+b\right)}+\frac{1+abc}{b\left(1+c\right)}+\frac{1+abc}{c\left(1+a\right)}\ge3\)
\(\Leftrightarrow\left[\frac{1+abc}{a\left(1+b\right)}+1\right]+\left[\frac{1+abc}{b\left(1+c\right)}+1\right]+\left[\frac{1+abc}{c\left(1+a\right)}+1\right]\ge6\)
\(\Leftrightarrow\frac{1+abc+ab+a}{a\left(1+b\right)}+\frac{1+abc+bc+b}{b\left(1+c\right)}+\frac{1+abc+c+ac}{c\left(1+a\right)}\ge6\)
\(\Leftrightarrow\frac{ab\left(c+1\right)+\left(a+1\right)}{a\left(1+b\right)}+\frac{bc\left(a+1\right)+\left(b+1\right)}{b\left(1+c\right)}+\frac{ac\left(b+1\right)+\left(c+1\right)}{c\left(1+a\right)}\ge6\)
\(\Leftrightarrow\frac{b\left(c+1\right)}{1+b}+\frac{a+1}{a\left(1+b\right)}+\frac{c\left(a+1\right)}{1+c}+\frac{b+1}{b\left(1+c\right)}+\frac{a\left(b+1\right)}{1+a}+\frac{c+1}{c\left(1+a\right)}\ge6\)
Ta có vế trái tương đương với:
\(\left[\frac{b\left(c+1\right)}{1+b}+\frac{b+1}{b\left(c+1\right)}\right]+\left[\frac{a\left(b+1\right)}{1+a}+\frac{1+a}{a\left(b+1\right)}\right]+\left[\frac{c\left(a+1\right)}{1+c}+\frac{1+c}{c\left(a+1\right)}\right]\)
\(\ge2+2+2=6\)
=> đpcm
