For all integers $n ≥ 2, n^3 > 2n + 1$

Question

I am having some serious trouble figuring out this induction problem. I've tried following other problems and can not seem to get the end result or understand it sufficiently.

My attempt:

Theorem: For all integers $n≥2, n^3 > 2n + 1$

Proof: We will prove this by induction. Let $P(n)$ be the statement: $n^3 > 2n + 1$. We will show $P(2)$ is true. When we let $n = 2, 2^3 = 8$ and $2(2) + 1 = 5$, so we know $P(2)$ to be true for $n^3 > 2n + 1$.

Induction Step: Suppose $P(k)$ is true for some integer $k≥2$, then $k^3>2(k)+1$. We wish to show that $P(k+1)$ is true, that is $(k+1)^3>2(k+1)+1$. We then factor it out to be $k^3 + 3k^2 + 3k + 1 > 2k + 3$. I am stuck from here. Any help would be great, thank you.

Answer 1

Here is the main part of the inductive step: \begin{align} (k+1)^3 &= k^3+3k^2+3k+1\tag{expand}\\[1em] &> (2k+1)+3k^2+3k+1\tag{by inductive hypothesis}\\[1em] &> 2k+2+1\tag{since $k\geq2$}\\[1em] &= 2(k+1)+1\tag{rearrange} \end{align}

Answer 2

To prove the inductive step, expand so that we have $$k^3 + 3k^2 + 3k + 1 > 2k + 3$$

By hypothesis, $k^3 > 2k+1$. It thus suffices to show $3k^2 + 3k + 1 > 2$, or, equivalently, $3(k + \frac{1}{2})^2 - \frac{7}{4} > 0$ which holds for $k\geq2$

Answer 3

You don't have to do it by induction. You can also look at $n^3-2n-1$.

$n^3-2n-1=n^3-n-n-1=n(n-1)(n+1)-(n+1)=(n+1)(n^2-n-1)=(n+1)(n^2-2n+1+n-2)=(n+1)(n-1)^2+(n+1)(n-2)>0$

We know from the condition $n\ge2$ that $n+1$ has to be positive, $(n-1)^2$ is always positive and so is $n-2$. So it's clear why the whole term has to be positive for $n\ge2$.

Answer 4

Here is a conceptual way of viewing the proof that makes it obvious, and works very generally. First construct the obvious inductive proof of the following Lemma

$\rm\:f(n) > 0\:$ for $\rm\:n\ge 2\ $ if $\rm\ f(2)> 0\:$ and $\rm\,f\,$ is increasing, i.e. $\rm\: f(n\!+\!1) \ge f(n)\:$ for $\rm\:n\ge2.\:$

So to show $\rm\:f(n) = n^3\!-2n-1> 0\:$ it suffices to show $\rm\: f(n\!+\!1)-f(n) = 3n^2\!+3n-1 \ge 0\:$ for $\rm\:n\ge 2,\:$ and $\,\rm f(2)>0,\,$ which is easy.

For many more examples see my posts on telescopy.