Alternative proof for $(m+1)^3 \leq2m^3$

Question

This is the proof on the textbook:

Until here, I was able to get to the same conclusion. What I tried to do was to prove by induction that $(m+1)^3 \leq2m^3$. I started from $(m + 2)^3$, but got nowhere.

The author then does the following:

I don't understand the first three lines of the previous picture (before the proof starts) nor how he uses it in the proof.

1. Is it possible to prove $(m+1)^3 \leq2m^3$ by induction? Or is there any other simpler proof that I'm not seeing?
2. Could someone please throw some light on the proof used in the textbook? I find it highly unintuitive.

Answer 1

The author made it too complicated.

$$(m+1)^3 \le 2m^3 \iff m+1 \le \sqrt[3]{2}m \iff m \ge \frac{1}{\sqrt[3]{2}-1}$$

Using a calculator you get $\frac{1}{\sqrt[3]{2}-1} \approx 3.8,$ but it's not difficult to prove it's between 3 and 4: $$ 3 < \frac{1}{\sqrt[3]{2}-1} < 4 \iff \frac 43 > \sqrt[3]{2} > \frac 54 \iff \frac{64}{27} > 2 > \frac{125}{64}.$$

So $(m+1)^3 \le 2m^3$, if $m\ge 4.$

Answer 2

1. Note that $(m+1)^3<2m^3\iff\left(\frac{m+1}m\right)^3<2\iff\left(1+\frac1m\right)^3<2$. This is easy to prove if $m=4$ and, if $m\geqslant4$, then $\left(1+\frac1m\right)^3\leqslant\left(1+\frac14\right)^3<2$.
2. The author is assuming that $k^3\leqslant2^k$. It follows from this that $2k^3\leqslant2^{k+1}$. So, in order to prove that $(k+1)^3\leqslant2^{k+1}$, it will be enough to prove that $(k+1)^3\leqslant2k^3$, which is equivalent to the assertion $3k^2+3k+1\leqslant k^3$.