<s>[INST] 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 $n$?
 [/INST] $n = 1 \implies a^5+1 - a^3-a = a^3(a^2 -1) - (a-1) = (a-1)(a^4+a^3-1)$. Observe that $a \to 1^{-} \implies a^4+a^3 - 1 \to 1 $, thus for $\epsilon = 0.01$ we can find a $\delta > 0$ such that $ -\delta < a - 1 < 0$, and $|a^4+a^3-1 - 1| < 0.01$ or $a^4+a^3-1 > 0.99$, thus $(a-1)(a^4+a^3-1) < 0$ on $(1-\delta, 1)$.
You did the case $n = 2$.
$n = 3$, $a^5+1- 2a^3 \ge  0$. For $0 < a \le 1\implies a^5 \ge a^6 \implies a^5+1 - 2a^3\ge a^6-2a^3 + 1 = (a^3-1)^2 \ge 0$, and for $\sqrt{\dfrac{6}{5}} > a  > 1, f(a) = a^5+1-2a^3\implies f'(a) = a^2(5a^2-6) < 0\implies f(a) < f(1) = 0$.
$n = 5$, choose $a > 1$ it does not hold.
$n = 4$. $a^5+1 - a^3 - a^4 = a^3(a^2-1) - (a^2-1)(a^2+1) = (a^2-1)(a^3-a^2-1)$. You can choose $a$ slightly bigger than $1$, then it fails to be greater than $0$. 
Thus $n = 2$ is the only value of $n$.
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