$5^a - a^2 \geq 5^{a-1}$ for $a \geq 12$?

Question

How do I show that $5^a - a^2 \geq 5^{a-1}$ for $a \geq 12$ where $a \in \mathbb{N}$? I'm trying to use this inequality as a part of larger proof but every time I read this statement while going through my proof, this jump seems a bit too abrupt(although I know it is true and can easily be checked via a calculator). I would be very grateful if someone could show me the steps in between and what I am missing in particular as this would also be teaching me what to do in the future when I encounter such a situation.

Answer 1

Rearrange to get $2^{a-1} \geq a^2$ for $a \geq 7$, then it's pretty easy to prove by induction.

Answer 2

We need to show that $2^a\geq2a^2$.

Indeed, for $a=7$ we obtain $128\geq98,$ which is true.

We'll prove that $$2^a\geq2a^2\Rightarrow2^{a+1}\geq2(a+1)^2.$$ Indeed, $$2^{a+1}\geq4a^2.$$ Thus, it's enough to prove that $$4a^2\geq2(a+1)^2$$ or $$a^2-2a-1\geq0,$$ which is true for $a\geq7$ and we are done by induction.