If $a$, $b$ and $c$ are sides of a triangle, then prove that $a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c)$ $\leqslant$ $3abc$

Using the Ravi substitution $$a=y+z,b=x+z,c=x+y$$ we have to prove that $$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2-6xyz\geq 0$$ But this is AM-GM:

$$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2\geq 6\sqrt[6]{x^6y^6z^6}=6xyz$$