Theo đề bài thì ta có:
\(\frac{1}{a+b+1}=1-\frac{1}{b+c+1}+1-\frac{1}{c+a+1}=\frac{b+c}{b+c+1}+\frac{c+a}{c+a+1}\)
\(\ge2.\sqrt{\frac{\left(b+c\right)\left(c+a\right)}{\left(b+c+1\right)\left(c+a+1\right)}}\left(1\right)\)
Tương tự ta có:
\(\hept{\begin{cases}\frac{1}{b+c+1}\ge2.\sqrt{\frac{\left(a+b\right)\left(c+a\right)}{\left(a+b+1\right)\left(c+a+1\right)}1}\left(2\right)\\\frac{1}{c+a+1}\ge2.\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\left(3\right)\end{cases}}\)
Nhân (1), (2), (3) vế theo vế ta được
\(\frac{1}{a+b+1}.\frac{1}{b+c+1}.\frac{1}{c+a+1}\ge8.\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\frac{1}{8}\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{4}\)
