Question

Cho ba so thực dương x,y,z thỏa mãn: \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge2\)

Tìm giá trị lớn nhất của biểu thức P=xyz

Answer 1

\(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge2\Rightarrow\frac{1}{1+x}\ge2-\frac{1}{1+y}\frac{1}{1+z}=\frac{y}{\left(1+y\right)}+\frac{z}{1+z}\ge2\sqrt{\frac{yz}{\left(1+y\right)\left(1+z\right)}};tt:\frac{1}{1+y}\ge2\sqrt{\frac{xz}{\left(1+x\right)\left(1+z\right)}};\frac{1}{1+z}\ge2\sqrt{\frac{xy}{\left(1+x\right)\left(1+y\right)}}\Rightarrow\frac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\frac{\left(xyz\right)^2}{\left(1+x\right)^2\left(1+y\right)^2\left(1+z\right)^2}}=\frac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\) \(\Rightarrow8xyz\le1\Rightarrow P_{max}=\frac{1}{8}\Leftrightarrow x=y=z=\frac{1}{2}\)