Question

Cho x, y, z là các số dương thỏa x + y + z = 1. CMR: \(\frac{\left(1-x\right)\left(1-y\right)\left(1-z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\le\frac{1}{8}\)

Answer 1

Đặt\(A=\frac{\left(1-x\right)\left(1-y\right)\left(1-z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)

\(A=\frac{\left(x+y\right)\left(y+z\right)\left(y+z\right)}{\left[\left(x+y\right)+\left(x+z\right)\right]\left[\left(x+y\right)+\left(y+z\right)\right]\left[\left(z+x\right)+\left(z+y\right)\right]}\)

Áp dụng BĐT AM-GM ta có:

\(A\le\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8.\sqrt{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}}=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{1}{8}\)

Dấu  " = " xảy ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)