Question

Cho ba số không âm a,b,c có tổng bằng 1. Chứng minh rằng:

\(4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le a+2b+c\)

 

Answer 1

Ta có:

\(4\left(1-a\right)\left(1-c\right)\left(1-b\right)\le4\left(1-b\right).\frac{\left(1-a+1-c\right)^2}{4}\)

\(=\left(1-b\right)\left(2-a-c\right)^2=\left(1-b\right)\left(a+2b+c\right)^2\)

\(=\left(1-b\right)\left(a+2b+c\right)\left(a+2b+c\right)\)

\(\le\left(a+2b+c\right).\frac{\left(a+2b+c+1-b\right)^2}{4}\)

\(=\left(a+2b+c\right).\frac{\left(a+b+c+1\right)^2}{4}\)

\(=\left(a+2b+c\right).\frac{4}{4}=a+2b+c\)

Dấu = xảy ra khi: 

\(\hept{\begin{cases}1-a=1-c\\a+2b+c=1-b\\a+b+c=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=c=\frac{1}{2}\\b=0\end{cases}}\)