Let $ a$, $ b$, $ c$ are sides of a triangle. Find the least possible value $ k$ such that the following inequality always holds:
$ \left|\frac{a\minus{}b}{a\plus{}b}\plus{}\frac{b\minus{}c}{b\plus{}c}\plus{}\frac{c\minus{}a}{c\plus{}a}\right|<k$
[i](Vitaly Lishunov)[/i]
1. Let \( a = x + y \), \( b = y + z \), and \( c = z + x \). These substitutions are valid because \( a, b, c \) are sides of a triangle, and any side of a triangle can be expressed as the sum of two other positive quantities.

2. We need to evaluate the expression:
   \[
   \left| \frac{a - b}{a + b} + \frac{b - c}{b + c} + \frac{c - a}{c + a} \right|
   \]
   Substituting \( a = x + y \), \( b = y + z \), and \( c = z + x \) into the expression, we get:
   \[
   \frac{(x + y) - (y + z)}{(x + y) + (y + z)} + \frac{(y + z) - (z + x)}{(y + z) + (z + x)} + \frac{(z + x) - (x + y)}{(z + x) + (x + y)}
   \]
   Simplifying each term:
   \[
   \frac{x - z}{2y + x + z} + \frac{y - x}{2z + y + x} + \frac{z - y}{2x + z + y}
   \]

3. To find the least possible value of \( k \) such that the inequality holds, we need to analyze the sum of these fractions. Notice that each fraction is of the form \(\frac{\text{difference}}{\text{sum}}\), and the sum of the denominators is always positive.

4. We can use the fact that the sum of the numerators is zero:
   \[
   (x - z) + (y - x) + (z - y) = 0
   \]
   This implies that the sum of the fractions is bounded by the maximum absolute value of each fraction.

5. By the triangle inequality, we know that:
   \[
   \left| \frac{x - z}{2y + x + z} \right| < 1, \quad \left| \frac{y - x}{2z + y + x} \right| < 1, \quad \left| \frac{z - y}{2x + z + y} \right| < 1
   \]
   Therefore, the sum of these fractions is less than 3 in absolute value.

6. To find the least possible value of \( k \), we need to consider the worst-case scenario where the sum of the absolute values of the fractions is maximized. Since each fraction is less than 1 in absolute value, the sum of the three fractions is less than 3.

7. Hence, the least possible value of \( k \) such that the inequality always holds is:
   \[
   k = 1
   \]

The final answer is \( \boxed{ k = 1 } \)