Question

CMR: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\) với a,b,c > 0

Answer 1

Đặt x = a+b , y = b+c , z = c+a

=> \(\begin{cases}a=\frac{x+z-y}{2}\\b=\frac{x+y-z}{2}\\c=\frac{y+z-x}{2}\end{cases}\)

Thay vào tính : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{x+z-y}{2y}+\frac{x+y-z}{2z}+\frac{y+z-x}{2x}\)

\(=\frac{1}{2}\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)\right]-\frac{3}{2}\) 

\(\ge\frac{1}{2}\left(2+2+2\right)-\frac{3}{2}=\frac{3}{2}\)

Answer 2