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 $n$?
 A: Note that equality always holds for $a = 1$.
$
a^5 + 1 \geq a^3 + a^n \\
\implies a^5 - a^3 \geq a^n - 1 \\
\implies a^3(a + 1)(a - 1) \geq a^n - 1
$
Case 1: $a > 1$. Then,
$$
a^4 + a^3 \geq 1 + a + a^2 + \cdots + a^{n-1}
$$
Case 2: $0 < a < 1$. Then,
$$
a^4 + a^3 \leq 1 + a + a^2 + \cdots + a^{n-1}
$$
Now, both cases hold true for $n = 2$ and certainly for $a = 1$.
Case 2 doesn't hold for $n=1$:
$$
a^4+a^3 \leq 1.
$$
Let $a = 1/x$ where $x>1$. Then, the inequality becomes $f(x) = x^4 - x - 1 \geq 0$. Now, $f(x)$ is a curve opening upwards which is negative at $x = 1$. Hence, $f(x)$ has a positive real root, say $\alpha > 1$. This means that $f(x) < 0$ in the interval $(1, \alpha)$, i.e. the inequality doesn't hold.
Case 1 doesn't hold for $n=3$:
$
a^4+a^3 \geq 1 + a + a^2 \\
\implies a^4 + a^3 + a^2 + a + 1 \geq 2(1 + a + a^2) \\
\implies a^5 - 1 \geq 2(a^3 - 1) \\
\implies a^5 - 2a^3 + 1 \geq 0.
$
Now, $p(a) = a^5 - 2a^3 + 1$ has a minima at $a_0 = \sqrt{6/5}$, where $p(a_0) = -0.05162... < 0$.
Case 1 doesn't hold for $n = 4$:
$
a^4 + a^3 > 1 + a + a^2 + a^3 \\
\implies q(a) = a^3 - a^2 - 1 \geq 0.
$
Now, $q(a) \rightarrow \infty$ as $a \rightarrow \infty$, and $q(1) = -1$. Hence, if a root of $q(a)$ be $a_0 > 1$, then $q(a)$ is negative in the interval $(1, a_0)$.
Case 1 clearly doesn't hold for $n>4$. Thus the inequality holds for the entire $\mathbb{R}^+$ only for $n=2$.
A: 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
$$a^5+1\ge a^3+a\;\;,\;\;\text{ false for}\;\;a=0.9 $$
A: $n = 1 \implies a^5+1 - a^3-a = a^3(a^2 -1) - (a-1) = (a-1)(a^4+a^3-1)$. 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 $n$.
