Prove that the inequality $\frac{n^3}{3} < 3n-3$ applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold

Question

I have been given the following task:

Prove that the following inequality applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold: $$\frac{n^3}{3} < 3n-3$$ My idea was to prove the statement by using induction.

For $n=1$ it follows:

$\frac{1^3}{3} = \frac{1}{3} \nless 0 = 3 * 1-3$

For $n=2$ it follows:

$\frac{2^3}{3} = \frac{8}{3} < 3 = 3 * 2-3$

For $n=3$ it follows:

$\frac{3^3}{3} = \frac{27}{3} = 9 \nless 6 = 3 * 3-3$

I now assume that the inequality does not hold for $n \in \mathbb{N} \backslash \{2\}.$

n $\rightarrow$ n+1

$\frac{(n+1)^3}{3} = \frac{n^3+3n^2+3n+1}{3} = \frac{n^3}{3}+ \frac{3n^2+3n+1}{3} < 3n-3 + \frac{3n^2+3n+1}{3}=\frac{3(3n-3)+3n^2+3n+1}{3}=\frac{3n^2+12n-8}{3}$

But how do I continue from this step or is this even the wrong approach?

Answer 1

Consider the opposite inequality:$$3n-3\leqslant\frac{n^3}3.$$As you proved, it holds for $n=3$. Now, assume that it holds for some $n\in\mathbb N\setminus\{1,2\}$. Then\begin{align}3(n+1)-3&=3n\\&=3n-3+3\\&\leqslant\frac{n^3}3+3\\&=\frac{n^3+9}3\\&\leqslant\frac{n^3+3n^2+3n+1}3\text{ (because $n>2$)}\\&=\frac{(n+1)^3}3.\end{align}

Answer 2

Let consider the reverse $\frac{n^3}{3} > 3n-3$ then we can prove the induction step as follows

$$\frac{(n+1)^3}{3} =\frac{n^3}{3}+\frac{3n^2+3n+1}{3}> 3n-3+\frac{3n^2+3n+1}{3}>3n-3+3=3(n+1)-3$$

we have used that for $n\ge 2$

$$\frac{3n^2+3n+1}{3}>3 \iff3n^2+3n-8>0$$

Answer 3

Show that $n^3/3 \ge 3n-3$ for $n\ge 3$, or equivalently

$n(n^2-9)\ge -9$, or

$n(n+3)(n-3) \ge -9$.

For $n \ge 3:$

$n(n+3)(n-3) \ge 0$ and we are done.