Question

Cho a,b,c là các số thực không âm thỏa mãn a+b+c=1. Tìm GTLN và GTNN của A=\(\frac{ab+bc+ac-abc}{a+2b+c}\)

Answer 1

\(A=\frac{b\left(a+c\right)+ac\left(1-b\right)}{\left(a+b\right)+\left(a+c\right)}\)

\(=\frac{b\left(1-b\right)+ac\left(1-b\right)}{\left(a+b\right)+\left(a+c\right)}\)

\(=\frac{\left(1-b\right)\left(ac+1-a-c\right)}{\left(a+b\right)+\left(a+c\right)}\)

\(=\frac{\left(1-b\right)\left(1-c\right)\left(1-a\right)}{\left(a+b\right)+\left(a+c\right)}\le\frac{\left[3-\left(a+b+c\right)\right]^3}{27\left(a+b\right)+\left(a+c\right)}\)

\(=\frac{8}{27}.\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{2}{27}.\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)