proof with induction $2^n + 4 > n^2 + 2n$ prove w/ full induction : $2^n + 4 > n^2 + 2n$.
Let's skip the other steps for a moment, I'm just getting stuck with the proof
the claim is: $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$
and now the proof  $$2^{n+1} + 4 > 2 * 2^n + 4$$ $$> 2 * (n^2 + 2n)$$ $$> 2n^2 + 4n$$
i am stuck here

It is true that for $n\geq5$
$$2^n+4>n^2+2n$$
Assume that for $n=k$
$$2^k+4>k^2+2k$$
Let's show for $n=k+1$
$$2^{k+1}+4>(k+1)^2+2(k+1)$$
Now
$$2(2^k+4)>2(k^2+2k)$$
$$2^{k+1}+4>2k^2+2k-4$$
$$2^{k+1}+4>2k^2+2k-4>(k+1)^2+2(k+1)$$
$$2k^2+2k-4>k^2+4k+3$$
$$k^2-2k-7>0$$
for $k\geq5$
$$(k-1)^2-8>0$$
