Question

Cho a,b,c là 3 số dương thỏa mãn a+b+c=1

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

Answer 1

Ta có:

\(VP=4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(\frac{1-a+1-c}{2}\right)^2\cdot4\left(1-b\right)=\left(2-a-c\right)\left(2-a-c\right)\left(1-b\right)\le\left(a+2b+c\right)\left(\frac{2-a-c+1-b}{2}\right)^2=\left(\frac{2+1-1}{2}\right)^2\left(a+2b+c\right)=VT\)

Đấu "=" khi \(\left\{{}\begin{matrix}a+b+c=1\\1-a=1-c\\2-a-c=1-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=c\\2a+b=1\\1-2a=-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1-2a=b\\1-2a=-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=0\\a=c=\frac{1}{2}\end{matrix}\right.\)