By the Cauchy-Schwarz inequality, we have
$$(a + b + c)(4a + 9b + 25c) \geq (2a + 3b + 5c)^2.$$
This implies $\frac{4a + 9b + 25c}{2a + 3b + 5c} \geq \frac{2a + 3b + 5c}{a + b + c}$. Adding similar inequalities, we obtain
$$\begin{aligned} & \frac{4a + 9b + 25c}{2a + 3b + 5c} + \frac{4b + 9c + 25a}{2b + 3c + 5a} + \frac{4c + 9a + 25b}{2c + 3a + 5b} \\ \geq & \frac{4a + 9b + 25c}{a + b + c} + \frac{4b + 9c + 25a}{b + c + a} + \frac{4c + 9a + 25b}{c + a + b} \\ = & \frac{10a + 10b + 10c}{a + b + c} \\ = & 10. \end{aligned}$$

By the given condition, the equality should hold. This means we need
$$a : b : c = 4a : 9b : 25c = 25a : 4b : 9c = 9a : 25b : 4c,$$
which means two of $a$, $b$, $c$ are zero, and the remaining can be any positive real number.