My question is about proving $r^3 \le 2$ by contradiction. I already proved (a) and half of (b)
Assume $r^3 > 2$
I want to define an h in such a way to show that $(r-h)^3 > 2$ to conclude that $r-h$ is an upper bound for E.
I start with $$(r-h)^3 = r^3 - 3r^2 h + 3rh^2 - h^3$$
Now I tried messing a lot to try to reach the inequality I need, but I do not know how to define $h$.
You have$$\begin{align}(r-h)^3>2&\iff r^3-3hr^2+3h^2r-h^3>2\\&\iff 3hr^2-3h^2r+h^3<r^3-2\\&\iff h(3r^2-3hr+h^2)<r^3-2.\end{align}$$Suppose that $0<h<1$. Then $3r^2-3hr+h^2<3r^2+h^2<3r^2+1$. So, take $h\in\left(0,\max\left\{1,\frac{r^3-2}{3r^2+1}\right\}\right)$, and then$$h(3r^2-3hr+h^2)<\frac{r^3-2}{3r^2+1}(3r^2+1)=r^3-2.$$