\(N=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+3\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\)
\(N\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}+3.\frac{9}{2\left(a+b+c\right)}=\frac{9}{6}+\frac{27}{6}=6\)
Dấu "=" khi \(a=b=c=1\)