Xét \(VT=a+2b+c=1+b\left(1\right)\)
Áp dụng BĐT AG-GM:
\(4\left(1-a\right)\left(1-c\right)\le\left(1-a+1-c\right)^2=\left(2-a-c\right)^2=\left(1+a+b+c-a-c\right)^2=\left(1+b\right)^2\left(2\right)\)
\(\Rightarrow4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(1-b\right)\left(1+b\right)^2\)
Mà \(\left(1-b\right)\left(1+b\right)^2-\left(1-b\right)=\left(1+b\right)\left(1-b^2-1\right)=-b^2\left(1+b\right)\le0,\forall b\ge0\)
Do đó \(\left(1-b\right)\left(1+b\right)^2\le1+b\left(3\right)\)
Từ \(\left(1\right)\left(2\right)\left(3\right)\) ta có ĐPCM
Dấu "=" \(\Leftrightarrow a=c=\dfrac{1}{2};b=0\)