Ta có: \(x+y+z=1\) nên:
\(\Rightarrow y+z=1-x\)
Thay \(y+z=1-x\) và áp dụng BĐT \(\left(a+b\right)^2\ge4ab\) ta được:
\(4\left(1-x\right)\left(1-y\right)\left(1-z\right)=4\left(y+z\right)\left(1-y\right)\left(1-z\right)\le\left[\left(y+z\right)+\left(1-z\right)\right]^2\left(1-y\right)\)
\(\Rightarrow4\left(y+z\right)\left(1-y\right)\left(1-z\right)\le\left(1+y\right)^2\left(1-y\right)=\left(1+y\right)\left(1-y^2\right)\le1+y\)
\(\Rightarrow4\left(1-x\right)\left(1-y\right)\left(1-z\right)\le1+y=x+2y+z\left(đpcm\right)\)
