Animated Solution for Mathematics - Sequence and Series: The minimum value of the sum of real numbers $a^{-5},a^{-4},3a^{-3},1,a^8$ and $a^{10}$ where $a>0$ is .........

We are tasked with finding the minimum value of the expression: where $a>0$.

To find the minimum value of a sum of positive terms, the Arithmetic Mean-Geometric Mean (AM-GM) inequality is our most powerful tool. It states that for positive real numbers $x_1,x_2,\dots ,x_n$:

The equality holds if and only if $x_1=x_2=\cdots =x_n$.

If we apply AM-GM directly to the given terms, we must consider the product of the terms. When multiplying powers of the same base, we add the exponents:

The product would be proportional to $a^6$. Because this product depends on $a$, the resulting minimum would not be a constant, rendering a direct application of AM-GM ineffective.

To obtain a constant minimum, the sum of the exponents must be zero. We can achieve this by splitting the term $3a^{-3}$ into three identical parts: $a^{-3}+a^{-3}+a^{-3}$.

Our new set of terms is $\{a^{-5},a^{-4},a^{-3},a^{-3},a^{-3},1,a^8,a^{10}\}$. Let us verify the sum of the exponents:

The variable $a$ now vanishes from the product, as $a^0=1$.

We now apply the AM-GM inequality to these $n=8$ terms:

Simplifying the product inside the root:

Multiplying both sides by $8$, we find:

The equality holds if and only if all terms are equal:

This condition is satisfied when $a=1$. Since $a=1$ is a valid positive real number, the minimum value is indeed achievable.

The minimum value of the expression is 8.