I am solving a proof by induction example. But I ended up with my hypothesis

$$
a\_{n-1} \geq \frac{2^n}{2}+n^2-2n+1
$$

and my inductive step

$$
a\_{n-1} \geq \frac{2^n}{2}+\frac{n^2}{2}-\frac{n}{2}.
$$

How can I show that if I am assuming that my hypothesis is true then my inductive step is also true?

It seems to me that I can compare both equation's right hand sides as

$$
\frac{2^n}{2}+n^2-2n+1 \geq \frac{2^n}{2}+\frac{n^2}{2}-\frac{n}{2} \Leftrightarrow -\frac{n^2}{2}-\frac{3}{2} n \geq -1 \Leftrightarrow -n^2-3n \geq -2.
$$

But now I just ended up with a neq inequality that I have to proof?

Edit
====

I am proving that $a\_n \geq 2^n + n^2$ for all natural numbers. $a\_n$ is defined by the recursive definition
$$
a\_n
=
\begin{cases}
1 & n=0 \\
n+2 a\_{n-1} & n \geq 1
\end{cases}
$$