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

Question

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 $>$.

Answer 1

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 $x < y < z$.

Answer 2

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

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

Answer 3

Hint:

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

Answer 4

Converting your expression so that all that is cubed is positive:

$$(z-x)^3 - (z-y)^3 - (y-x)^3$$ $$(A+B)^3-A^3-B^3$$

with $A,B$ positive. Since cubing is concave up with $0^3=0$, this expression must be positive.

Answer 5

Hint, $(z-x)^3=\left((z-y)+(y-x)\right)^3$.