Since $|a+b| \le |a|+|b|$, $|b+c| \le |b|+|c|$ and $|a+c| \le |a|+|c|$, we have $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|} \le\frac{|a|+|b|+|b|+|c|+|a|+|c|}{|a|+|b|+|c|}=2$

The minimum is $\frac{2}{3}$, which is attained for $a = b = 1$, $c = -1$. Let $a$, $b$, and $c$ be arbitrary real numbers, not all of them equal $0$. By flipping signs, we can assume that at least two of $a$, $b$, and $c$ are non-negative. Actually, without loss of generality, we can assume that $a, b\geq 0$. Then, \begin{align*} 3|a + b| + 3|b + c| + 3|c + a| &\geq 3a + 3b + 3|b+c| + 3|a+c| \\ &\geq 2(a+b) + (a + |a + c|) + (b + |b + c|) \\ &= 2(|a| + |b|) + (|-a| + |a + c|) + (|-b| + |b + c|) \\ &\geq 2(|a| + |b|) + |c| + |c| \\ &= 2|a| + 2|b| + 2|c| \end{align*} We have proved that the minimum possible value of $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|}$ is $\frac{2}{3}$.