Determine the maximum value of $m$, such that the inequality

\[ (a^2+4(b^2+c^2))(b^2+4(a^2+c^2))(c^2+4(a^2+b^2)) \ge m \]

holds for every $a,b,c \in \mathbb{R} \setminus \{0\}$ with $\left|\frac{1}{a}\right|+\left|\frac{1}{b}\right|+\left|\frac{1}{c}\right|\le 3$.
When does equality occur?
1. **Assume \(a, b, c \geq 0\)**: If the inequality holds for \(a, b, c \geq 0\), it will also be true for all \(a, b, c \in \mathbb{R}\). Therefore, we can assume without loss of generality that \(a, b, c \geq 0\).

2. **Plugging in \(a = b = c = 1\)**: 
   \[
   (1^2 + 4(1^2 + 1^2))(1^2 + 4(1^2 + 1^2))(1^2 + 4(1^2 + 1^2)) = (1 + 8)(1 + 8)(1 + 8) = 9^3 = 729
   \]
   This suggests that \(m \leq 729\).

3. **Using QM-HM and GM-HM inequalities**:
   \[
   \sqrt[3]{\prod_{\text{sym}} \sqrt{\frac{a^2 + 4b^2 + 4c^2}{9}}} \geq \sqrt[3]{\prod_{\text{sym}} \frac{9}{\frac{1}{a} + \frac{4}{b} + \frac{4}{c}}} \geq \frac{3}{\sum_{\text{sym}} \frac{\frac{1}{a} + \frac{4}{b} + \frac{4}{c}}{9}} = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \geq 1
   \]
   Let's break this down step-by-step:
   - By the QM-HM inequality, we have:
     \[
     \sqrt{\frac{a^2 + 4b^2 + 4c^2}{9}} \geq \frac{3}{\frac{1}{a} + \frac{4}{b} + \frac{4}{c}}
     \]
   - Taking the product of these inequalities for \(a, b, c\):
     \[
     \prod_{\text{sym}} \sqrt{\frac{a^2 + 4b^2 + 4c^2}{9}} \geq \prod_{\text{sym}} \frac{3}{\frac{1}{a} + \frac{4}{b} + \frac{4}{c}}
     \]
   - Taking the cube root of both sides:
     \[
     \sqrt[3]{\prod_{\text{sym}} \sqrt{\frac{a^2 + 4b^2 + 4c^2}{9}}} \geq \sqrt[3]{\prod_{\text{sym}} \frac{3}{\frac{1}{a} + \frac{4}{b} + \frac{4}{c}}}
     \]
   - Simplifying the right-hand side:
     \[
     \sqrt[3]{\prod_{\text{sym}} \frac{3}{\frac{1}{a} + \frac{4}{b} + \frac{4}{c}}} = \frac{3}{\sum_{\text{sym}} \frac{\frac{1}{a} + \frac{4}{b} + \frac{4}{c}}{9}} = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}
     \]
   - Given the condition \(\left|\frac{1}{a}\right| + \left|\frac{1}{b}\right| + \left|\frac{1}{c}\right| \leq 3\), we have:
     \[
     \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \geq 1
     \]

4. **Conclusion**: 
   \[
   \sqrt[3]{\prod_{\text{sym}} \sqrt{\frac{a^2 + 4b^2 + 4c^2}{9}}} \geq 1
   \]
   Therefore, the best constant \(m\) is indeed \(9^3 = 729\).

5. **Equality condition**: Equality occurs if \((a, b, c) = (\pm 1, \pm 1, \pm 1)\).

The final answer is \(\boxed{729}\)