Áp dụng BĐT AM-GM ta có:

\(\dfrac{1}{a+1}\ge1-\dfrac{1}{b+1}+1-\dfrac{1}{c+1}+1-\dfrac{1}{d+1}\)

\(=\dfrac{b}{b+1}+\dfrac{c}{c+1}+\dfrac{d}{d+1}\)\(\ge3\sqrt[3]{\dfrac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}\)

Tương tự cho 3 BĐT còn lại cũng có:

\(\dfrac{1}{1+b}\ge3\sqrt[3]{\dfrac{acd}{\left(a+1\right)\left(c+1\right)\left(d+1\right)}};\dfrac{1}{c+1}\ge3\sqrt[3]{\dfrac{abd}{\left(a+1\right)\left(b+1\right)\left(d+1\right)}};\dfrac{1}{d+1}\ge3\sqrt[3]{\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

Nhân theo vế 4 BĐT trên ta có:

\(\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\ge81\sqrt[3]{\left(\dfrac{abcd}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\right)^3}\)

\(\Leftrightarrow1\ge81abcd\Leftrightarrow abcd\le\dfrac{1}{81}\)