P = $\frac{3}{b+c-a}$ + $\frac{4}{a+c-b}$ + $\frac{5}{a+b-c}$ = ($\frac{1}{b+c-a}$ + $\frac{1}{a+c-b}$ + 2($\frac{1}{b+c-a}$ + $\frac{1}{a+b-c}$ ) + 3($\frac{1}{a+c-b}$ + $\frac{1}{a+b-c}$ $\geq$ $\frac{2}{c}$ + $\frac{4}{b}$ + $\frac{6}{c}$= $\frac{6}{a}$ + 2a $\geq$ 4$\sqrt[n]{3}$ dấu = xảy ra khi a=b=c=$\sqrt[n]{3}$