Question

cho a,b,c>0 và a+b+c=1. tìm min F= \(\frac{a}{1+b-a}\)+\(\frac{b}{1+c-b}\)+\(\frac{c}{1+a-c}\)

Answer 1

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:\(F=\frac{a}{1+b-a}+\frac{b}{1+c-b}+\frac{c}{1+a-c}\)

\(=\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\)

\(=\frac{a^2}{2ab+ac}+\frac{b^2}{2bc+ab}+\frac{c^2}{2ac+bc}\)

\(\ge\frac{\left(a+b+c\right)^2}{2ab+ac+2bc+ab+2ac+bc}=\frac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)

\(\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\) khi \(a=b=c=\frac{1}{3}\)