Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}=2\)
\(\Leftrightarrow\dfrac{1}{a+b+1}=1-\dfrac{1}{b+c+1}+1-\dfrac{1}{c+a+1}\)
\(=\dfrac{b+c}{b+c+1}+\dfrac{c+a}{c+a+1}\ge2\sqrt{\dfrac{\left(b+c\right)\left(c+a\right)}{\left(b+c+1\right)\left(c+a+1\right)}}\)
Tương tự cho 2 BĐT còn lại rồi nhân theo vế:
\(\dfrac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\ge\dfrac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\)
\(1\ge8\left(a+b\right)\left(b+c\right)\left(c+a\right)\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{8}\)
