Proof of $3 \leq \frac{4 a+b}{a+4 b}+\frac{4 b+c}{b+4 c}+\frac{4 c+a}{c+4 a}$:

1. Variant: Multiplying by the main denominator and simplifying leads to the equivalent inequality $45 a b c \leq 7\left(a^{2} b+b^{2} c+c^{2} a\right)+8\left(a b^{2}+b c^{2}+c a^{2}\right)$. This follows from the inequality between the arithmetic and geometric mean (e.g., $a^{2} b+b^{2} c+c^{2} a \geq 3 \sqrt[3]{a^{2} b \cdot b^{2} c \cdot c^{2} a}=3 a b c$).
2. Variant: For $n \geq 1$ and real numbers $a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n}$, the Cauchy-Schwarz inequality $\sqrt{a_{1}^{2}+\ldots+a_{n}^{2}} \sqrt{b_{1}^{2}+\ldots+b_{n}^{2}} \geq a_{1} b_{1}+\ldots+a_{n} b_{n}$ holds. For $n=3, a_{1}^{2}=\frac{4 a+b}{a+4 b}, a_{2}^{2}=\frac{4 b+c}{b+4 c}, a_{3}^{2}=\frac{4 c+a}{c+4 a}$, $b_{1}^{2}=(4 a+b)(a+4 b), b_{2}^{2}=(4 b+c)(b+4 c), b_{3}^{2}=(4 c+a)(c+4 a)$, we get

$$\left(\frac{4 a+b}{a+4 b}+\frac{4 b+c}{b+4 c}+\frac{4 c+a}{c+4 a}\right)((4 a+b)(a+4 b)+(4 b+c)(b+4 c)+(4 c+a)(c+4 a)) \geq(5(a+b+c))^{2}$$

The second factor is less than $\frac{25}{3}(a+b+c)^{2}$ because $2(a+b+c)^{2}-6(a b+b c+c a)=(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$, from which the claim follows.
Proof of $\frac{4 a+b}{a+4 b}+\frac{4 b+c}{b+4 c}+\frac{4 c+a}{c+4 a} > \frac{1}{4}$; The latter holds because

$$\frac{b}{a+4 b}+\frac{c}{b+4 c}+\frac{a}{c+4 a}>\frac{b}{4 a+4 b+4 c}+\frac{c}{4 a+4 b+4 c}+\frac{a}{4 a+4 b+4 c}=\frac{1}{4}$$

Remarks: 1. The inequality is invariant under cyclic permutation of $a$ to $b$, $b$ to $c$, and $c$ to $a$, but not under arbitrary permutation of $a, b, c$. Therefore, one can assume without loss of generality that $a \geq b$ and $a \geq c$, but not $a \geq b \geq c$. Many solution approaches only worked under the latter assumption.
2. Some participants examined the limit for $a \rightarrow \infty$ with fixed $b$ and $c$ and estimated $\frac{4 b+c}{b+4 c}$. This is not a proof of the right inequality, as the sum for finite values of $a$ can be larger than for $a \rightarrow \infty$.
3. The number $\frac{33}{4}$ can be approximated arbitrarily closely for $a=1, b=\varepsilon, c=\varepsilon^{2}$ as $\varepsilon \rightarrow 0$.