Solution:
Note,
$$\begin{array}{r l r} \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)} = \frac{\left(\frac{1}{a}\right)^{2}}{a(b+c)} + \frac{\left(\frac{1}{b}\right)^{2}}{b(a+c)} + \frac{\left(\frac{1}{c}\right)^{2}}{c(a+b)} \\ & & \geq \frac{\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^{2}}{2(ab + ac + bc)} \\ & & = \frac{\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^{2}}{2\left(\frac{1}{c} + \frac{1}{b} + \frac{1}{a}\right)} \\ & & = \frac{\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)}{2} \\ & & \geq \frac{3}{2\sqrt[3]{a b c}} \\ & & = \frac{3}{2} \end{array} \quad \text{(AM-GM)}$$
Now, $(a + b + c)^{2} \geq 3(ab + ac + bc)$ for positive reals $a, b, c$. $(\dagger)$
So we also get,
$$\begin{array}{r l r} \frac{1}{a^{2}(b+c)} + \frac{1}{b^{2}(a+c)} + \frac{1}{c^{2}(a+b)} \geq \frac{\left(\frac{1}{a}\right)^{2}}{(b+c)} + \frac{\left(\frac{1}{b}\right)^{2}}{(a+c)} + \frac{\left(\frac{1}{c}\right)^{2}}{(a+b)} \\ & & \geq \frac{\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^{2}}{2(a + b + c)} \\ & & = \frac{\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^{2}}{2\left(\frac{1}{b} \cdot \frac{1}{c} + \frac{1}{a} \cdot \frac{1}{c} + \frac{1}{a} \cdot \frac{1}{b}\right)} \\ & & = \frac{3}{2} \end{array} \quad (\dagger)$$
Finally, as $a^{2} + 1 \geq 2a$, we get
$$\frac{a^{2} + 2025}{a^{3}(b + c)} + \frac{b^{2} + 2025}{b^{3}(a + c)} + \frac{c^{2} + 2025}{c^{3}(a + b)} \geq \frac{2a + 2024}{a^{3}(b + c)} + \frac{2b + 2024}{b^{3}(a + c)} + \frac{2c + 2024}{c^{3}(a + b)}$$
$$\qquad \geq 2 \cdot \frac{3}{2} + 2024 \cdot \frac{3}{2}$$
$$\qquad = 3039$$
Equality can be achieved when $a = b = c = 1$.