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 $n$?

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 $n=2$.
