<s>[INST] 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 $>$.
 [/INST] 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
$3(d_1 d_2)(d_1 + d_2) > 0$.
</s>