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 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}
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 $k\geq2$