$\frac{3(a + b + c)^2}{25(ab + bc + ca)(a^2 + b^2 + c^2)} \le \sum_{cyc}\frac{1}{(3a + b + c)(c + 4a)} \le \frac{\sqrt{3(ab + bc + ca)}}{25abc}$ 
Given postives $a, b, c$. Prove that $$\frac{3(a + b + c)^2}{25(ab + bc + ca)(a^2 + b^2 + c^2)} \le \sum_{cyc}\frac{1}{(3a + b + c)(c + 4a)} \le \frac{\sqrt{3(ab + bc + ca)}}{25abc}$$

This is not an overkill. Sure it is definitely not without the Buffalo Way and a computer borrowed from NASA or one used to study quantum computing.
One particularly out-of-the-ordinary (not unintelligible) way to solve this inequality is to let
$$\begin{cases}
a + b + c = 3\\
ab + bc + ca = 3(1 - m)\\
a^2 + b^2 + c^2 = 3(1 + 2m)
\end{cases} (0 \le m < 1)$$
(, which leads to nothing. This is not my idea at all.)
To prove the inequality between the first and second expression, we have that $$\sum_{cyc}\frac{1}{(3a + b + c)(c + 4a)} = \sum_{cyc}\frac{\dfrac{b^2}{c + 4a}}{b^2(3a + b + c)} \ge \frac{\displaystyle \sum_{cyc}\left(\dfrac{b}{\sqrt{c + 4a}}\right)^2}{\displaystyle \sum_{cyc}b^2(3a + b + c)}$$
$$ \ge \frac{\displaystyle \sum_{cyc}\left(\dfrac{b}{\sqrt{c + 4a}}\right)^2 \cdot \sum_{cyc}b(c + 4a)}{\displaystyle \left[(a^2 + b^2 + c^2)(a + b + c) + 2 \cdot \sum_{cyc}c^2a\right] \cdot \sum_{cyc}b(c + 4a)}$$
$$ \ge \frac{(a + b + c)^3}{\dfrac{5}{3}(a^2 + b^2 + c^2)(a + b + c) \cdot 5(ab + bc + ca)} = \frac{3(a + b + c)^2}{25(ab + bc + ca)(a^2 + b^2 + c^2)}$$
But I can't prove the one between the second and third. Help me.

For the proof of the right inequality we can use BW.
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that:
$$3(ab+ac+bc)\geq625a^2b^2c^2\left(\sum_{cyc}\frac{1}{(3a+b+c)(c+4a)}\right)^2,$$ which is something obvious: 
https://www.wolframalpha.com/input/?i=3%28xy%2Bxz%2Byz%29-625x%5E2y%5E2z%5E2%281%2F%28%283x%2By%2Bz%29%284x%2Bz%29%29%2B1%2F%28%283y%2Bx%2Bz%29%284y%2Bx%29%29%2B1%2F%28%283z%2Bx%2By%29%284z%2By%29%29%29%5E2%2Cx%3Da%2Cy%3Da%2Bu%2Cz%3Da%2Bv
