Prove that if $a^5-a^3+a=3$, then $a^6\geq 5$ The problem, which I encountered in a highschool book, goes as following:

Prove that if $a^5-a^3+a=3,$ then $a^6\geq 5$ must hold.

Now, obviously, I have tried a lot of things such as:
$a^6=a^4-a^2+3a\Rightarrow a^4-a^2+3a\geq 5$ or multiplying that again by $a^2$ to create a new expression for $a^6$ and simplify some terms, but higher powers keep appearing and it just doesnt seem to end

The equation $a^5-a^3+1-3=0$ has at least one real solution.
As
$$3=a(a^4-a^2+1)=a\bigl((a^2-1/2)^2+3/4\bigr)$$
we have $a>0$ assuming $a$ is real.
Now
$$a^4-a^2=\frac3a-1,$$
hence
$$a^6=a^4-a^2+3a=\frac3a+3a-1
=3(\sqrt{a}-1/\sqrt{a})^2+6-1.$$
