Minimum Value of Sum of Positive Powers of a Real Number

To find the minimum value of the sum of the real numbers $a^{-5}$, $a^{-4}$, $3a^{-3}$, $1$, $a^8$, and $a^{10}$ for $a > 0$, we can express the given sum $S$ as:

$S = a^{-5} + a^{-4} + 3a^{-3} + 1 + a^8 + a^{10}$

We can split the term $3a^{-3}$ into three equal terms $a^{-3} + a^{-3} + a^{-3}$, giving a total of $8$ positive terms:

$S = a^{-5} + a^{-4} + a^{-3} + a^{-3} + a^{-3} + 1 + a^8 + a^{10}$

Since $a > 0$, all 8 terms are strictly positive. We can apply the Arithmetic Mean - Geometric Mean (AM-GM) inequality to these 8 positive terms:

$\frac{a^{-5} + a^{-4} + a^{-3} + a^{-3} + a^{-3} + 1 + a^8 + a^{10}}{8} \ge \left( a^{-5} \cdot a^{-4} \cdot a^{-3} \cdot a^{-3} \cdot a^{-3} \cdot 1 \cdot a^8 \cdot a^{10} \right)^{1/8}$

Simplifying the term inside the 8th root by adding the exponents of $a$:

$-5 + (-4) + (-3) + (-3) + (-3) + 0 + 8 + 10 = -18 + 18 = 0$

Therefore, the product simplifies to:

$\left( a^0 \right)^{1/8} = 1^{1/8} = 1$

Substituting this back into the inequality gives:

$\frac{S}{8} \ge 1 \implies S \ge 8$

The equality holds if and only if all the terms involved in the AM-GM inequality are equal:

$a^{-5} = a^{-4} = a^{-3} = 1 = a^8 = a^{10}$

This condition is satisfied when $a = 1$, which is a valid positive real number ($a > 0$).

Thus, the minimum value of the given sum is $8$.