Proof that for any $r^3 > 2$ there exists an $h>0$ such that $(r-h)^3>2$ I am trying to prove that for any $r^3 > 2$ there exists an $h>0$ such that $(r-h)^3>2$. I know this is true and I'm trying to prove it using only high school level algebra. So no intermediate value theorems or continuity of real numbers.
So far I have this
$$(r-h)^3=(r-h)(r^2-2rh+h^2\ )=r^3-3r^2\ h+3rh^2-h^3$$
$$r^3-\left(r-h\right)^3=3r^2h-3rh^2+h^3$$
Since $h^2>0$
$$<3r^2h+h^3$$
However, I don't know how to proceed. I know for $r$'s close to $2$ then $h<1$. But I can't find an expression that assures $h>0$ and that simplifies to at most $r^3-2$ (so as to guarantee that $r^3-\left(r-h\right)^3<r^3-2$ which implies $(r-h)^3>2$).
Thank you for your help!
 A: $r^3 > 2\implies r>\sqrt[3]2$. Show that exists an $h>0$ such that $(r-h)^3>2$.  $(r-h)^3>2 \implies r-h > \sqrt[3] 2 \implies -h>\sqrt[3] 2 -r \implies h<r-\sqrt[3] 2  .$Note that $h>0 $ since $r>\sqrt[3]2$. Just take $0<h<r-\sqrt[3]2$. $\square$
A: For $0 < h < r$ is
$$
 (r-h)^3 = r^3 - 3r^2h + \underbrace{3rh^2 - h^3}_{> 0} > r^3 - 3r^2 h
$$
so that the choice
$$
 h = \frac{r^3-2}{3r^2} > 0 
$$
gives
$$
 (r-h)^3 > r^3 - 3r^2 h = 2 \, .
$$
This also happens to be one step from Newton's method applied to the function $f(x) = x^3 - 2$.
A: we are allowed to make restrictions on the variables. Since $1^3 = 1$   we know $r > 1. $  Next, as a separate case we point out that, when $r \geq 2,$  then $r^3 \geq 8,  r^3 - 2 \geq 6;$ but then $\left(\frac{3}{2}\right)^3 - 2 = \frac{11}{8} < 6$   does better.
We continue with $1 < r < 2 .$  Note that
$$h^2 - 3hr + 3 r^2  = \left( h - \frac{3r}{2}  \right)^2  + \frac{3}{4}  r^2 \geq 0 $$   and is strictly positive  unless $h=r=0$
As $1 < r < 2$  and  we want  $ 1 < r-h <2,$   we may demand $0 < h < 1.$
With these restrictions,
$$h^2 - 3hr + 3 r^2  < h^2 + 3 r^2  < 1 + 12 = 13  $$
for any $(r,h)$
You wanted   $ r^3 - 2  >  h(h^2 - 3hr + 3 r^2).$  Just take
$$   h =  \frac{r^3 - 2}{13}.$$
