Find a costant $C$, such that $$   \frac{S}{ab+bc+ca}\le  C$$ where $a,b,c$ are the side lengths of an arbitrary triangle, and $S$ is the area of the triangle. 
(The maximal number of points is given for the best possible constant, with proof.)
To find the best possible constant \( C \) such that 
\[ \frac{S}{ab + bc + ca} \leq C \]
where \( a, b, c \) are the side lengths of an arbitrary triangle and \( S \) is the area of the triangle, we proceed as follows:

1. **Express the area \( S \) using Heron's formula:**
   The area \( S \) of a triangle with side lengths \( a, b, c \) can be expressed using Heron's formula:
   \[
   S = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   where \( s \) is the semi-perimeter of the triangle:
   \[
   s = \frac{a + b + c}{2}
   \]

2. **Use the formula for the area of a triangle in terms of its sides and the circumradius \( R \):**
   Another formula for the area \( S \) of a triangle is:
   \[
   S = \frac{abc}{4R}
   \]
   where \( R \) is the circumradius of the triangle.

3. **Relate \( S \) to \( ab + bc + ca \):**
   We need to find a relationship between \( S \) and \( ab + bc + ca \). Using the formula for the area in terms of the circumradius, we have:
   \[
   S = \frac{abc}{4R}
   \]
   We also know that for any triangle, the following inequality holds:
   \[
   ab + bc + ca \geq 4\sqrt{3}S
   \]
   This inequality can be derived from the fact that the maximum value of \( \frac{S}{ab + bc + ca} \) occurs when the triangle is equilateral.

4. **Consider the case of an equilateral triangle:**
   For an equilateral triangle with side length \( a \), the area \( S \) is:
   \[
   S = \frac{\sqrt{3}}{4}a^2
   \]
   and the sum of the products of the sides is:
   \[
   ab + bc + ca = 3a^2
   \]
   Therefore, for an equilateral triangle:
   \[
   \frac{S}{ab + bc + ca} = \frac{\frac{\sqrt{3}}{4}a^2}{3a^2} = \frac{\sqrt{3}}{12}
   \]

5. **Determine the best possible constant \( C \):**
   From the above steps, we see that the best possible constant \( C \) is:
   \[
   C = \frac{\sqrt{3}}{12} = \frac{1}{4\sqrt{3}}
   \]

Thus, the best possible constant \( C \) such that \( \frac{S}{ab + bc + ca} \leq C \) for any triangle is \( \frac{1}{4\sqrt{3}} \).

The final answer is \( \boxed{\frac{1}{4\sqrt{3}}} \).