\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)
\(=\left(\frac{a}{b+c}+\frac{b+c}{a}\right)+\left(\frac{b}{a+c}+\frac{a+c}{b}\right)+\left(\frac{c}{a+b}+\frac{a+b}{c}\right)\ge2+2+2=6\)
vậy min \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=6\)