Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{x+1}\ge1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}\)\(=\dfrac{y}{y+1}+\dfrac{z}{z+1}\)
\(\ge2\sqrt{\dfrac{yz}{\left(y+1\right)\left(z+1\right)}}\). Tương tự ta cũng có:
\(\dfrac{1}{y+1}\ge2\sqrt{\dfrac{xz}{\left(x+1\right)\left(z+1\right)}};\dfrac{1}{z+1}\ge2\sqrt{\dfrac{xy}{\left(x+1\right)\left(y+1\right)}}\)
Nhân theo vế 3 BĐT trên ta có:
\(\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge8\sqrt{\dfrac{\left(xyz\right)^2}{\left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2}}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\dfrac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(\Leftrightarrow1\ge8xyz\Leftrightarrow xyz\le\dfrac{1}{8}\)
Xảy ra khi \(x=y=z=\dfrac{1}{2}\)
