Áp dụng BĐT AM-GM ta có: \(\frac{1}{x+1}=\)\(1-\frac{1}{y+1}+1-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{ỹz}{\left(y+1\right)\left(z+1\right)}}\)
Tương tự ta cũng có: \(\frac{1}{y+1}\ge2\sqrt{\frac{xz}{\left(x+1\right)\left(z+1\right)}};\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)
Nhân từng vế của ba BĐT trên ta được:
\(\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge8\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\Rightarrow xyz\le\frac{1}{8}\)