For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$? For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$?

1.$1$
2.$2$
3.$3$
4.$4$
5.$5$

My attempt:It's clear that it is true for $n=2$.Because:
$(a^3-1)(a^2-1) \ge 0$
Is true because $a^3-1$ and $a^2-1$ are both negative or both positive.But there is a problem:How can we make sure it doesn't hold for any other $n$?

Clearly $\;n=5\;$ is out of question, as $\;a^3\le1\;$ isn't true for all $\;a>0\;$.
Something similar with $\;n=4\;$ , as then 
$$a^5+1\ge a^3+a^4\iff (a+1)(a^4-a^3+a^2-a+1)\ge a^3(a+1)$$
which is false for values close to $\;1\;$ from the right, for example $\;a=1.1\;$ (check this).
With $\;n=3\;$ we'd get
$$a^5+1\ge 2a^3\;,\;\;\text{ false, again for}\;\;a=1.1$$
Finally, for $\;n=1\;$ we'd get
$$a^5+1\ge a^3+a\;\;,\;\;\text{ false for}\;\;a=0.9 $$
