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: 
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.](https://math.stackexchange.com/search?q=user:242+telescopy)
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 $k\geq2$