An Algebraic Proof that $|y^3 - x^3| \ge |(y - x)|^3/4 $ I can prove this using calculus, but not by simple algebra: can anyone help ?

Calculus Proof:
Fix the separation of $x$ and $y$ so that $y = x + d$ with $d>0$ ($ \implies y > x \implies y^3 > x^3$) and now consider $ f(x) = y^3 - x^3 = (x+d)^3 - x^3$.
$ f(x) = (x+d)^3 - x^3$ = $3x^2d + 3d^2x +d^3$ which is a quadratic in $x$ with a minimum given by $f`(x) = 0 = 6xd + 3d^2$ giving $x = - d/2$ and therefore $y = d/2$. This is not surprising considering the geometry: it says that for a fixed separation of $x$ and $y$, $y^3 - x^3$ is minimised when $x$ and $y$ are symetrically placed around the inflexion point of the cubic.
So, $y^3 - x^3 = d^3/8 - (-d^3)/8 = d^3/4$ and since this is the minimum value it follows that for $x \in (-\infty, +\infty)$ and $y > x$ then $y^3 - x^3 \ge (y - x)^3/4 $, so that  $|y^3 - x^3| \ge |(y - x)|^3/4 $ whether $y > x$ or $y < x$ (and clearly this is true for $y = x$.)

(The original interest in the inequality comes from this question: cauchy sequence on $\mathbb{R}$)

Wlog. $y\ge x$. Then 
$$\begin{align}y^3-x^3&\ge \frac{(y-x)^3}{4}&\iff\\
4(y^3-x^3)-(y-x)^3&\ge 0&\iff\\
3y^3+3y^2x-3x^3-3x^2y&\ge 0&\iff\\
3(y+x)^2(y-x)&\ge0\end{align}$$
