Question

CM BĐT: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) với \(a,b,c>0\) 

Answer 1

\(VT=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)

\(=\dfrac{1}{2}\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)-3>=\dfrac{9}{2}-3=\dfrac{3}{2}\)