Question

Cho \(x,y,z\ge0\)T/M : \(x+y+z=1\)

CMR : \(x+2y+z\ge4\left(1-x\right)\left(1-y\right)\left(1-z\right)\)

Answer 1

Áp dụng BĐT AM - GM, ta có:

\(4\left(1-x\right)\left(1-y\right)\left(1-z\right)\)

\(=4\left(y+z\right)\left(x+z\right)\left(x+y\right)\)

\(\le\frac{\left(x+y+y+z\right)^2}{4}\times4\left(x+z\right)\)

\(=\left(x+2y+z\right)^2\left(x+z\right)\)

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

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

\(=x+2y+z\left(\text{đ}pcm\right)\)

Dấu "=" xảy ra khi a = b = c = \(\frac{1}{3}\)

Answer 2

Dấu = xảy ra:\(\hept{\begin{cases}x=z=\frac{1}{2}\\y=0\end{cases}}\)