prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$ 
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$

Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then I would get $\geq$ instead of $>$.

Notice that the polynomial
$$S = (x-y)^3 + (y-z)^3 + (z-x)^3$$
is zero whenever two of the variables are equal so we must have
$$S = C(y-x) (z-x) (z-y)$$
for some constant $C$. A direct computation gives $C=3$ and the result follows since all factors are positive when $x < y < z$.
