We need to minimize $S=a^{-5}+a^{-4}+3a^{-3}+1+a^8+a^{10}, \,\, a>0.$

We will rewrite the expression in a form where AM-GM can be applied neatly.

1. Group the terms

Notice that $a^{-5}+a^{10}$ and $a^{-4}+a^8$ are natural pairs, because their exponents can be balanced.

Also, $3a^{-3}=a^{-3}+a^{-3}+a^{-3}.$

So, $S=(a^{-5}+a^{10})+(a^{-4}+a^8)+(a^{-3}+a^{-3}+a^{-3})+1.$

2. Apply AM-GM to each pair

For positive numbers, AM-GM gives $x+y\ge 2\sqrt{xy}.$

Pair 1:

$a^{-5}+a^{10}\ge 2\sqrt{a^{-5}\cdot a^{10}}=2\sqrt{a^5}=2a^{5/2}.$

Pair 2:

$a^{-4}+a^8\ge 2\sqrt{a^{-4}\cdot a^8}=2\sqrt{a^4}=2a^2.$

Thus, $S\ge 2a^{5/2}+2a^2+3a^{-3}+1.$

This is true, but not immediately enough to get a clean minimum. So let us instead look for a point where all terms become simple and check whether AM-GM can directly produce a constant lower bound.

3. Test the natural equality point $a=1$

At $a=1$, $S=1+1+3+1+1+1=8.$

So the minimum is at most $8$.

Now we prove that $S\ge 8$ for all $a>0$.

4. Prove the lower bound $S\ge 8$

Using AM-GM on suitable grouped terms:

First group:

$a^{-5}+a^{-4}+a^{-3}\ge 3\sqrt[3]{a^{-5}a^{-4}a^{-3}}=3a^{-4}.$

This alone is not enough. A better grouping is to compare each term with 1 through AM-GM:

For positive $x$, $x+1\ge 2\sqrt{x}.$

Apply to $a^{-5}$ and $a^{10}$: $a^{-5}+a^{10}\ge 2\sqrt{a^5}=2a^{5/2}.$

Apply to $a^{-4}$ and $a^8$: $a^{-4}+a^8\ge 2\sqrt{a^4}=2a^2.$

So $S\ge 2a^{5/2}+2a^2+3a^{-3}+1.$

Now at $a=1$, this becomes $8$, and the equality conditions of both AM-GM inequalities require $a^{-5}=a^{10} \Rightarrow a^{15}=1 \Rightarrow a=1,$ $a^{-4}=a^8 \Rightarrow a^{12}=1 \Rightarrow a=1.$

Hence $a=1$ is the common equality point.

To ensure it is indeed the minimum, define $f(a)=a^{-5}+a^{-4}+3a^{-3}+1+a^8+a^{10}.$

Differentiate: $f'(a)=-5a^{-6}-4a^{-5}-9a^{-4}+8a^7+10a^9.$

At $a=1$, $f'(1)=-5-4-9+8+10=0.$

Second derivative: $f''(a)=30a^{-7}+20a^{-6}+36a^{-5}+56a^6+90a^8>0 \quad (a>0).$

So $f$ is strictly convex on $(0,\infty)$, and the critical point $a=1$ is the unique global minimum.

Therefore, $\boxed{8}$

5. Compare with stored answer

Stored correct answer: $8$

Our derived answer is also $8$, so they agree.