12. Given that $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are all positive numbers, determine the largest real number $C$ such that the inequality $C\left(x_{1}^{2005}+x_{2}^{2005}+x_{3}^{2005}+x_{4}^{2005}+x_{5}^{2005}\right) \geqslant x_{1} x_{2} x_{3} x_{4} x_{5}\left(x_{1}^{125}+x_{2}^{125}+x_{3}^{125}+x_{4}^{125}+x_{5}^{125}\right)^{16}$
always holds. (2005 Brazilian Mathematical Olympiad problem)
12 . From Example 9, we get
$$\begin{array}{l}
5\left(x_{\mathrm{L}}^{2005}+x_{2}^{2005}+x_{3}^{2005}+x_{4}^{2005}+x_{5}^{2005}\right) \geqslant \\
\left(x_{1}^{5}+x_{2}^{5}+x_{3}^{5}+x_{4}^{5}+x_{5}^{5}\right)\left(x_{1}^{2000}+x_{2}^{2000}+x_{3}^{2000}+x_{4}^{2000}+x_{5}^{2000}\right)
\end{array}$$

By the AM-GM inequality, we have
$$x_{\mathrm{E}}^{5}+x_{2}^{5}+x_{3}^{5}+x_{4}^{5}+x_{5}^{5} \geqslant 5 x_{1} x_{2} x_{3} x_{4} x_{5}$$

By the Power Mean inequality, we have
$$\frac{x_{1}^{2000}+x_{2}^{2000}+x_{3}^{2000}+x_{4}^{2000}+x_{5}^{2000}}{5} \geqslant\left(\frac{x_{1}^{125}+x_{2}^{125}+x_{3}^{125}+x_{4}^{125}+x_{5}^{125}}{5}\right)^{16}$$

Therefore, the maximum value of \( C \) such that \( C\left(x_{1}^{2005}+x_{2}^{2005}+x_{3}^{2005}+x_{4}^{2005}+x_{5}^{2005}\right) \geqslant x_{1} x_{2} x_{3} x_{4} x_{5}\left(x_{1}^{125}+x_{2}^{125}+\right. \left.+x_{4}^{125}+x_{5}^{125}\right)^{16} \) holds is \( 5^{15} \).