Ta có : \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
=> \(1:\frac{a}{b+c}=1:\frac{b}{a+c}=1:\frac{c}{a+b}\)
=> \(\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{b+c+a+c+a+b}{a+b+c}=2\)
Khi đó : P = \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{b+c}{a}.3=2.3=6\)