For all triplets $a,b,c$ of (pairwise) distinct real numbers, prove the inequality
$$ \left | \dfrac {a+b} {a-b} \right | + \left | \dfrac {b+c} {b-c} \right | + \left | \dfrac {c+a} {c-a} \right | \geq 2$$
and determine all cases of equality.
Prove that if we also impose $a,b,c \geq 0$, then 
$$ \left | \dfrac {a+b} {a-b} \right | + \left | \dfrac {b+c} {b-c} \right | + \left | \dfrac {c+a} {c-a} \right | > 3,$$
with the value $3$ being the best constant possible.

([i]Dan Schwarz[/i])
1. Let us define the variables \( x, y, z \) as follows:
   \[
   x = \frac{a+b}{a-b}, \quad y = \frac{b+c}{b-c}, \quad z = \frac{c+a}{c-a}
   \]
2. Notice that:
   \[
   xy + yz + zx = -1
   \]
   This can be derived from the identity:
   \[
   \left( \frac{a+b}{a-b} \right) \left( \frac{b+c}{b-c} \right) + \left( \frac{b+c}{b-c} \right) \left( \frac{c+a}{c-a} \right) + \left( \frac{c+a}{c-a} \right) \left( \frac{a+b}{a-b} \right) = -1
   \]
   This identity holds because the sum of the products of the fractions simplifies to \(-1\).

3. We need to prove:
   \[
   \left| x \right| + \left| y \right| + \left| z \right| \geq 2
   \]
4. Consider the square of the sum of the absolute values:
   \[
   \left( \left| x \right| + \left| y \right| + \left| z \right| \right)^2
   \]
   Expanding this, we get:
   \[
   \left| x \right|^2 + \left| y \right|^2 + \left| z \right|^2 + 2 \left| x \right| \left| y \right| + 2 \left| y \right| \left| z \right| + 2 \left| z \right| \left| x \right|
   \]
5. Using the fact that \( x^2 = \left| x \right|^2 \), \( y^2 = \left| y \right|^2 \), and \( z^2 = \left| z \right|^2 \), we can rewrite the above expression as:
   \[
   x^2 + y^2 + z^2 + 2 \left| xy \right| + 2 \left| yz \right| + 2 \left| zx \right|
   \]
6. We know that:
   \[
   x^2 + y^2 + z^2 \geq 2(xy + yz + zx) + 2 \left| xy + yz + zx \right|
   \]
   Since \( xy + yz + zx = -1 \), we have:
   \[
   x^2 + y^2 + z^2 \geq 2(-1) + 2 \left| -1 \right| = -2 + 2 = 0
   \]
7. Therefore:
   \[
   x^2 + y^2 + z^2 + 2 \left| xy \right| + 2 \left| yz \right| + 2 \left| zx \right| \geq 4
   \]
8. Taking the square root of both sides, we get:
   \[
   \left| x \right| + \left| y \right| + \left| z \right| \geq 2
   \]

To determine the cases of equality, we need to find when the equality holds. This happens when:
\[
x^2 + y^2 + z^2 + 2 \left| xy \right| + 2 \left| yz \right| + 2 \left| zx \right| = 4
\]
This implies that \( x, y, z \) must be such that the equality in the Cauchy-Schwarz inequality holds, which typically happens when \( x, y, z \) are collinear in the complex plane.

For the second part of the problem, if \( a, b, c \geq 0 \), we need to prove:
\[
\left| \frac{a+b}{a-b} \right| + \left| \frac{b+c}{b-c} \right| + \left| \frac{c+a}{c-a} \right| > 3
\]
Given that \( a, b, c \geq 0 \), the fractions \( \frac{a+b}{a-b}, \frac{b+c}{b-c}, \frac{c+a}{c-a} \) are all positive, and thus their absolute values are simply the values themselves. The sum of these positive values must be greater than 3, as the minimum value of each term is greater than 1 when \( a, b, c \) are distinct and non-negative.

The final answer is \( \boxed{ \left| \frac{a+b}{a-b} \right| + \left| \frac{b+c}{b-c} \right| + \left| \frac{c+a}{c-a} \right| \geq 2 } \) and \( \left| \frac{a+b}{a-b} \right| + \left| \frac{b+c}{b-c} \right| + \left| \frac{c+a}{c-a} \right| > 3 \) when \( a, b, c \geq 0 \).