Xét giả thiết : \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge2\Leftrightarrow\frac{1}{1+x}\ge\left(1-\frac{1}{1+y}\right)+\left(1-\frac{1}{1+z}\right)\)
\(\Leftrightarrow\frac{1}{1+x}\ge\frac{y}{1+y}+\frac{z}{1+z}\ge2\sqrt{\frac{yz}{\left(1+y\right)\left(1+z\right)}}\)
Tương tự : \(\frac{1}{1+y}\ge2\sqrt{\frac{xz}{\left(1+x\right)\left(1+z\right)}}\) ; \(\frac{1}{1+z}\ge2\sqrt{\frac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Nhân các bđt trên theo vế : \(\frac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\frac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
\(\Rightarrow1\ge8xyz\Rightarrow xyz\le\frac{1}{8}\)
Dấu "=" xảy ra khi \(\begin{cases}\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=2\\\frac{1}{1+x}=\frac{1}{1+y}=\frac{1}{1+z}\end{cases}\) \(\Leftrightarrow x=y=z=\frac{1}{2}\)
Vậy max (xyz) = 1/8 <=> x = y = z = 1/2
