Question

cho a,b,c>00 và a+b+c=1. cm:\(\frac{c+ab}{a+b}+\frac{a+bc}{b+c}+\frac{b+ca}{c+a}\ge2\)

Answer 1

\(VT=\sum\frac{c\left(a+b+c\right)+ab}{a+b}=\sum\frac{\left(a+c\right)\left(b+c\right)}{a+b}\)

Đặt \(\left(a+b;b+c;c+a\right)=\left(x;y;z\right)\Rightarrow x+y+z=2\)

\(VT=\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}=\frac{1}{2}\left[\left(\frac{xy}{z}+\frac{yz}{x}\right)+\left(\frac{xy}{z}+\frac{zx}{y}\right)+\left(\frac{yz}{x}+\frac{zx}{y}\right)\right]\)

\(VT\ge\frac{1}{2}\left(2\sqrt{\frac{xy^2z}{xz}}+2\sqrt{\frac{x^2yz}{yz}}+2\sqrt{\frac{xyz^2}{xy}}\right)=x+y+z=2\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)