The minimum value of the sum of real number `a^(-5),a^(-4),3a^(-3),1,a^8,a n da^(10)w i t ha >0` is

Explanation

To find the minimum value of the sum \( S = a^{-5} + a^{-4} + 3a^{-3} + 1 + a^{-8} + a^{-10} \) for \( a > 0 \), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Rewrite the Expression**:
The given expression can be rewritten as:
\[ S = a^{-5} + a^{-4} + a^{-3} + a^{-3} + a^{-3} + 1 + a^{-8} + a^{-10} \]

2. **Count the Terms**:
We have a total of 8 terms in the expression:
\[ a^{-5}, a^{-4}, a^{-3}, a^{-3}, a^{-3}, 1, a^{-8}, a^{-10} \]

3. **Apply the AM-GM Inequality**:
According to the AM-GM inequality, the arithmetic mean of non-negative numbers is greater than or equal to their geometric mean. Thus, we can write:
\[ \frac{a^{-5} + a^{-4} + a^{-3} + a^{-3} + a^{-3} + 1 + a^{-8} + a^{-10}}{8} \geq \sqrt[8]{a^{-5} \cdot a^{-4} \cdot a^{-3} \cdot a^{-3} \cdot a^{-3} \cdot 1 \cdot a^{-8} \cdot a^{-10}} \]

4. **Calculate the Geometric Mean**:
The product of the terms in the geometric mean is:
\[ a^{-5} \cdot a^{-4} \cdot a^{-3} \cdot a^{-3} \cdot a^{-3} \cdot 1 \cdot a^{-8} \cdot a^{-10} = a^{-5 - 4 - 3 - 3 - 3 - 8 - 10} = a^{-38} \]
Therefore, the geometric mean is:
\[ \sqrt[8]{a^{-38}} = a^{-\frac{38}{8}} = a^{-\frac{19}{4}} \]

5. **Set Up the Inequality**:
Now substituting back into the AM-GM inequality gives:
\[ \frac{S}{8} \geq a^{-\frac{19}{4}} \]

6. **Find the Minimum Value**:
To minimize \( S \), we need to maximize \( a^{\frac{19}{4}} \). The minimum occurs when all terms are equal, which leads to:
\[ S \geq 8 \cdot a^{-\frac{19}{4}} \]
Setting \( a^{-\frac{19}{4}} = 1 \) gives \( a^{\frac{19}{4}} = 1 \), or \( a = 1 \).

7. **Calculate \( S \) at \( a = 1 \)**:
Substituting \( a = 1 \) into \( S \):
\[ S = 1^{-5} + 1^{-4} + 3 \cdot 1^{-3} + 1 + 1^{-8} + 1^{-10} = 1 + 1 + 3 + 1 + 1 + 1 = 8 \]

8. **Conclusion**:
Therefore, the minimum value of \( S \) is:
\[ \boxed{8} \]