Question

Cho các số x,y,z ko âm thoả mãn x+y+z=1.CMR

x+2y+z\(\ge\)4(1-x)(1-y)(1-z)

Answer 1

Đặt: \(\left\{{}\begin{matrix}x+y=a\\y+z=b\\x+z=c\end{matrix}\right.\Leftrightarrow a+b+c=2\)

\(bđt\Leftrightarrow a+b\ge4abc\Leftrightarrow4\left(a+b\right)\ge16abc\)

Mà: \(4\left(a+b\right)=\left(a+b\right)\left(a+b+c\right)^2=\left(a+b\right)\left[\left(a+b\right)+c\right]^2\ge4\left(a+b\right)^2c\ge16abc\) (bđt \(\left(m+n\right)^2\ge4mn\))

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}a=b\\a+b=c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=y+z\\x+2y+z=x+z\end{matrix}\right.\Leftrightarrow x=z=\frac{1}{2};y=0\)