Determine the smallest real number $C$ such that the inequality \[ C(x_1^{2005} +x_2^{2005} + \cdots + x_5^{2005}) \geq x_1x_2x_3x_4x_5(x_1^{125} + x_2^{125}+ \cdots + x_5^{125})^{16}  \] holds for all positive real numbers $x_1,x_2,x_3,x_4,x_5$.
1. **Identify the inequality and the goal:**
   We need to find the smallest real number \( C \) such that the inequality
   \[
   C(x_1^{2005} + x_2^{2005} + \cdots + x_5^{2005}) \geq x_1 x_2 x_3 x_4 x_5 (x_1^{125} + x_2^{125} + \cdots + x_5^{125})^{16}
   \]
   holds for all positive real numbers \( x_1, x_2, x_3, x_4, x_5 \).

2. **Apply Muirhead's inequality:**
   Muirhead's inequality states that for any symmetric sum of monomials, if the sequence of exponents of one sum majorizes the sequence of exponents of another, then the inequality holds. 

3. **Rewrite the inequality in terms of symmetric sums:**
   We need to compare the sequences of exponents on both sides of the inequality. The left-hand side is:
   \[
   C(x_1^{2005} + x_2^{2005} + \cdots + x_5^{2005})
   \]
   The right-hand side is:
   \[
   x_1 x_2 x_3 x_4 x_5 (x_1^{125} + x_2^{125} + \cdots + x_5^{125})^{16}
   \]

4. **Analyze the exponents:**
   - The left-hand side has the exponents \((2005, 0, 0, 0, 0)\) for each term.
   - The right-hand side can be expanded as follows:
     \[
     x_1 x_2 x_3 x_4 x_5 (x_1^{125} + x_2^{125} + \cdots + x_5^{125})^{16}
     \]
     Each term in the expansion of \((x_1^{125} + x_2^{125} + \cdots + x_5^{125})^{16}\) will have exponents that sum to \(2000\) (since \(125 \times 16 = 2000\)). Multiplying by \(x_1 x_2 x_3 x_4 x_5\) adds 1 to each exponent, resulting in terms with exponents summing to \(2005\).

5. **Determine the majorization:**
   The sequence \((2005, 0, 0, 0, 0)\) must majorize the sequence of exponents on the right-hand side. The right-hand side will have terms like \((125k_1 + 1, 125k_2 + 1, \ldots, 125k_5 + 1)\) where \(k_1 + k_2 + k_3 + k_4 + k_5 = 16\).

6. **Find the smallest \( C \):**
   By Muirhead's inequality, the smallest \( C \) that satisfies the inequality is when the sequences are equal. The sum of the exponents on the right-hand side is \(2005\), and the left-hand side is already \(2005\). Therefore, the smallest \( C \) is determined by the number of terms, which is \(5^{15}\).

   Thus, the smallest \( C \) is:
   \[
   C = 5^{15}
   \]

The final answer is \( \boxed{ 5^{15} } \)