Stuck while trying to prove $2k^3 \geq (k + 1)^3$...... how can I prove the following:
$2k^3 \geq (k + 1)^3$
This is the final part of the elaborate proof for $2^n > n^3 $ give $ n \geq 10$
I have used induction and end up with:
$ 2^{K+1} > 2k^3 $
After that we just need to prove $2k^3 \geq (k + 1)^3$, but I have no idea how to do so. :(
So how do I proceed to prove $2k^3 \geq (k + 1)^3$? 
OR
Am I trying to prove $ 2^{K+1} > 2k^3 $ in a wrong manner?
Thanks for all the help in advance! :D
 A: The result is equivalent to $((k+1)/k)^3\leq 2$ so it is equivalent to $(1+1/k)^3\leq 2$.
But the sequence $(1+1/n)^3$ decreases, so if it is $\leq 2$ in $10$ it will be also for $n\geq 10$.
A: In order to prove $2k^3 \geq (k + 1)^3$ we can expand the inequality to get: $k^3 \geq 3k^2+3k+1$. Theorem: If $[a,b]$ is some interval, $f(a)<g(a)$, and $f'(x)≤g'(x)$ on the interval, then $f(x)<g(x)$ on the interval. In our case the interval is $I:= [10,\infty)$.  $2x \geq 3$ on $I$ and $10^2>2(10)+1$ so $x^2>2x+1$ and $3x^2>6x+3$  on $I$. $10^3>3(10)^2+3(10)+1$ and we know that $3x^2>6x+3$ so we can conclude that $x^3 \geq 3x^2+3x+1$. Note that I used the theorem in a sort of backwards way: I chose my inequalities based on what I need to prove. Note: My previous answer had a subtle flaw which was pointed out to me by @awllower.
A: By subtracting and expanding, the inequality is equivalent with $$k^3\ge 3k^2+3k+1,$$ or $$k(k^2-3k-3)\ge1.$$
When $k\ge10,$ the factor $(k^2-3k-3)$ is always greater than $1$ (in fact, this holds when $k\ge4.$) So the inequality holds.
Hope this helps.
A: If you make the cubic root of both members you get
$(\sqrt[3]{2}-1) k \geqslant 1$ that holds for $k \geqslant 4$
I hope I gave the right answer to the question
