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I'm new here and unsure if this is the right way to format a problem, but here goes nothing. I'm currently trying to solve an inequality proof to show that $n^3 > 2n+1$ for all $n \geq 2$.

I proved the first step $(P(2))$, which comes out to $8>5$, which is true.

In the next step we assume that for some $k \geq 2$, $k^3 > 2k+1$. Then we consider the quotient $\frac{f(k+1)}{f(k)} > \frac{g(k+1)}{g(k)}$.

I so far have simplified it to the following:

$$\begin{align*} \frac{(k+1)^3}{k^3} &> \frac{2(k+1)+1}{2k+1}\\ &= \frac{k^3+3k^2+3k+1)}{k^3}\\ &> \frac{2k+3}{2k+1}\\ &= 1 + \frac{3}{k} + \frac{3}{k^2} + \frac{1}{k^3}\\ &> \frac{2k+3}{2k+1} \end{align*}$$

I don't know how to simplify the right side anymore (my algebra is terrible). I know that I have to simplify that inequality and multiply it by our previous assumption.

I should end up with some variant of $(k+1)^3 > 2(k+1)+1$. (This is $P(k+1)$). I just need help simplifying. Thanks!

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  • $\begingroup$ Actually, it may be better to modify the inequality so that you are comparing to $0$: $n^3-2n-1\gt 0$. Then you can apply subtraction and verify that step $n+1$ is greater than step $n$. $\endgroup$ – abiessu Sep 28 '13 at 17:06
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Hint: For the inductive step you can assume that $k^3-2k-1>0$ for some $k \in \mathbb{N}$ such that $k \ge 2$, then for $k+1$

$$(k+1)^3-2(k+1)-1= \cdots >0 $$

(in $\cdots$ you should use the hypothesis, $k^3-2k-1>0$) because for $k \ge 2$, $3k(k+1)\ge 2$ so $3k(k+1) -1\ge 1>0$.

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Alternative Approach (for fun!):

Note that $n^3 > n^2$ for $n \geq 2$.

Check your original inequality explicitly for $n = 2$:

$n^3 = 8 > 5 = 2n+1$, good.

Next, when is it true that $n^2 > 2n+1$?

In other words, when is it true that $n^2 - 2n - 1 > 0$?

This is just a parabola that opens upwards, so you can find the vertex.

Using calculus (or the $n = -b/2a$ "fact") we find the vertex is at $(1, -2)$.

Thus, the graph of the values are strictly increasing for $n \geq 1$.

We see that at $n = 3$ we get $n^2 - 2n - 1 = 2$, at which point it is already positive.

Thus, $n^3 > n^2 > 2n+1$ for $n \geq 3$, and we checked the initial inequality explicitly for $n = 2$.

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