I am having some trouble with this proof.
Part of it is that I have to prove it for every integer. Does this mean I have an inductive step that goes for $P(k+1)$ and $P(k-1)$? Assuming my base case is $n=0$.
Prove $n^2 \geq n$ for every integer.
Starting with the base case $n=0$:
We assume $P(k)$ is true for $k=n$.
Want to show $P(k+1)$ holds, that is
$(k+1)^2 \geq k+1$
Case 1: $P(k+1)$, for integers $\geq 0$
$k^2 \geq k$
$k^2+1 \geq k+1$
$(k^2+1)^2 \geq (k+1)^2$
$k^4+2k^2+1 \geq k^2+2k+1$
$k^2(k^2+1)+(k^2+1) \geq (k^2+k)+(k+1)$