Show that $n^3 > (n+1)^2$ for $n>2$ using mathematical induction

Is the following a correct way of showing that $n^3 > (n+1)^2$ for $n>2$ using mathematical induction? Thank you in advance!

$P(n): n^3>(n+1)^2$

$P(3): 3^3>(3+1)^2$

$27 > 16$ (correct)

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

Assume $P(k)$ is true.

$P(k)$ $\implies$ $P(k+1)$?

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

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

$(3k-2)(k+1)>0$ $\forall$ $k\in N$

$\implies 3k^2+k-2>0$

$\implies 4k^2+5k+2>k^2+4k+4$

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

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

Hence $P(k) \implies P(k+1)$

Hence by the principal of mathematical induction $P(n)$ is true $\forall n>2$

• What is it in all these student induction proofs with the statement "assume $P(k)$ is true" at the start of the inductive step? I know it's logical nitpicking, but it's still incorrect. I'm guessing a lot of HS teachers don't understand the nuance. Nov 27 '14 at 0:08

That looks fine, but here is a simpler way to do the inductive step: assuming $k\ge3$ and $k^3>(k+1)^2$, we have \eqalign{(k+1)^3 &=k^3+3k^2+3k+1\cr &>(k+1)^2+3+2k+0\cr &=k^2+4k+4\cr &=(k+2)^2\ .\cr}