prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$

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 $x < y < z$.

Because it is only the differences in the variables that matter, we can subtract the smallest value ($x$) from each term:

$-y^3 + (y-z)^3 + z^3 > 0$.

Let $y = d_1$ and $z-y = d_2$ and note $z = d_1 + d_2$. Then:

$-d_1^3 - d_2^3 + (d_1 + d_2)^3 > 0$.

Simplify:

$3d_1 d_2^2 + 3 d_1^2 d_2 >0$.

or

$3(d_1 d_2)(d_1 + d_2) > 0$.

Hint:

If $a+ b +c=0, a^3+b^3+c^3=3abc$