Question

[Dưa Hấu]

Có x;y;z >0 và  \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=2\)

CMR: \(xyz\le\frac{1}{8}\)

Answer 1

[Đinh Đức Hùng]

Từ (gt) \(\Rightarrow\frac{1}{1+x}=\left(1-\frac{1}{1+y}\right)+\left(1-\frac{1}{1+z}\right)=\frac{y}{1+y}+\frac{z}{1+z}\ge2\sqrt{\frac{yz}{\left(1+y\right)\left(1+z\right)}}\)

Tương tự \(\hept{\begin{cases}\frac{1}{1+y}\ge2\sqrt{\frac{xz}{\left(1+x\right)\left(1+z\right)}}\\\frac{1}{1+z}\ge2\sqrt{\frac{xy}{\left(1+x\right)\left(1+y\right)}}\end{cases}}\)

\(\Rightarrow\frac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\frac{\left(xyz\right)^2}{\left[\left(1+x\right)\left(1+y\right)\left(1+z\right)\right]^2}}=\frac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)

\(\Rightarrow xyz\le\frac{1}{8}\)