IMO 1964 - Problem 2

Problem: Denote by $a,b,c$ the lengths of the sides of a triangle. Prove that $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$.
Solution: By substituting $a=x+y,b=y+z,c=z+x(x,y,z>0)$ the given inequality becomes $6xyz\le x^2y+xy^2+y^2z+yz^2+z^2x+zx^2$, which follows immediately by the AM-GM inequality applied to $x^2y,xy^2,y^z,yz^2,z^x,zx^2$.