Proving the following inequality using Mathematical Induction

Question

I need to prove the following inequality is true for n > 2.

$n^3$ > 2$n^2$ + 3

1. Prove using base case n = 3:

$3^3$ > 2$(3)^2$ + 3

27 > 2(9) + 3

27 > 18 + 3

27 > 21 (true)

2. Assume true for n = k:

$k^3$ > 2$k^2$ + 3

3. Prove for n = k +1

$(k+1)^3$ > 2$(k+1)^2$ + 3

I'm not sure where to go from here. I'm thinking I may need to prove it for n = k +2 (since I'm proving the expression is true for n > 2), but I'm not sure. How would I be able to prove this using n = k + 1?

Answer 1

Note that\begin{align}(k+1)^3&=k^3+3k^2+3k+1\\&>2k^2+3+3k^2+3k+1\text{ (by the assumption)}\\&=5k^2+3k+4.\end{align}So, all that you have to prove now is that $5k^2+3k+4\geqslant2(k+1)^2+3=2k^2+4k+5$. In other words, what remains to be proved is that $k^2\geqslant k+1$, which is quite easy.

Answer 2

Now, $$(k+1)^3=k^3+3k^2+3k+1=2(k+1)^2+3+k^3+k^2-k-4>2(k+1)^2+3.$$

Answer 3

Hint: $(k+1)^3>2(k+1)^2+3$ is equal to $-k^3>-k^2+k+4$ now we have $$k^3>2k^2+3>-k^2+k+4$$ if $$k^2-k+7>0$$ which is true for $k\geq 3$