# Showing $2^x$ is a lower bound

How do I show that $$2^x - x^2 \in \Omega(2^x)$$?

Basically, I know that this means that $$\exists a, x_0 \in \mathbb{R^+}, \forall x \in \mathbb{N}, a.2^x \leq 2^x - x^2$$.

I worked around a bit with this and I seemed to have gotten something that makes sense for the case of an upper bound; $$2^x -x^2 < 2^x + x^2 < 2^{x+1} = 2.2^n$$ Now, if I were trying to find an upper bound, I'd have chosen $$a = 2$$ and $$x_0 = 1$$. How do I work the other way around(basically less than or equal to $$2^x - x^2$$ and prove this lower bound?

• Show that by mathematical induction, that $$2x^2< 2^x$$ for $$x>7$$.
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