Proof that for any $r^3 > 2$ there exists an $h>0$ such that $(r-h)^3>2$

Question

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!

Answer 1

$r^3 > 2\implies r>\sqrt[3]2$. 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 $0<h<r-\sqrt[3]2$. $\square$

Answer 2

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 $f(x) = x^3 - 2$.

Answer 3

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 $$ h = \frac{r^3 - 2}{13}.$$