Question

[dbrby]

cho x,y,z ≥ 0 thỏa mãn x + y + z = 2. Cmr: \(x+2y+z\ge\left(2-x\right)\left(2-y\right)\left(2-z\right)\)

Answer 1

[Nguyễn Việt Lâm]

\(VP=\left(2-x\right)\left(2-z\right)\left(2-y\right)=\left(y+z\right)\left(x+y\right)\left(2-y\right)\le\frac{\left(x+2y+z\right)^2}{4}\left(2-y\right)\)

\(VP\le\left(x+2y+z\right).\frac{\left(x+2y+z\right)\left(2-y\right)}{4}\le\left(x+2y+z\right)\frac{\left(x+y+z+2\right)^2}{16}=x+2y+z\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=z=1\\y=0\end{matrix}\right.\)