Determine the smallest and the greatest possible values of the expression 
$$\left( \frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right)\left( \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)$$ provided $a,b$ and $c$ are non-negative real numbers satisfying $ab+bc+ca=1$.

[i]Proposed by Walther Janous, Austria [/i]
1. **Define the expression and simplify:**
   Let 
   \[
   S = \frac{1}{a^2+1} + \frac{1}{b^2+1} + \frac{1}{c^2+1}
   \]
   We need to find the smallest and greatest possible values of 
   \[
   \left( \frac{1}{a^2+1} + \frac{1}{b^2+1} + \frac{1}{c^2+1} \right) \left( \frac{a^2}{a^2+1} + \frac{b^2}{b^2+1} + \frac{c^2}{c^2+1} \right)
   \]
   Notice that 
   \[
   \frac{a^2}{a^2+1} = 1 - \frac{1}{a^2+1}
   \]
   Therefore, the expression can be rewritten as:
   \[
   \left( \frac{1}{a^2+1} + \frac{1}{b^2+1} + \frac{1}{c^2+1} \right) \left( 3 - \left( \frac{1}{a^2+1} + \frac{1}{b^2+1} + \frac{1}{c^2+1} \right) \right)
   \]
   Let \( S = \frac{1}{a^2+1} + \frac{1}{b^2+1} + \frac{1}{c^2+1} \). Then the expression becomes:
   \[
   S(3 - S) = 3S - S^2
   \]

2. **Establish bounds for \( S \):**
   We need to find the bounds for \( S \). We claim that:
   \[
   \frac{9}{4} \geq S \geq 2
   \]
   **Proof of lower bound:**
   \[
   S = \frac{1}{a^2+1} + \frac{1}{b^2+1} + \frac{1}{c^2+1}
   \]
   Using the condition \( ab + bc + ca = 1 \), we can transform \( S \) as follows:
   \[
   S = \frac{1}{a^2 + ab + bc + ca} + \frac{1}{b^2 + ab + bc + ca} + \frac{1}{c^2 + ab + bc + ca}
   \]
   Since \( ab + bc + ca = 1 \), we have:
   \[
   S = \frac{1}{(a+b)(a+c)} + \frac{1}{(b+c)(b+a)} + \frac{1}{(c+a)(c+b)}
   \]
   Using the AM-GM inequality:
   \[
   \frac{1}{a^2+1} \geq \frac{1}{2a}
   \]
   Summing up, we get:
   \[
   S \geq \frac{2(a+b+c)}{(a+b)(b+c)(c+a)}
   \]
   We need to show:
   \[
   \frac{2(a+b+c)}{(a+b)(b+c)(c+a)} \geq 2
   \]
   Simplifying:
   \[
   2(a+b+c) \geq 2((a+b+c)(ab+bc+ca) - abc)
   \]
   \[
   2(a+b+c) \geq 2(a+b+c) - 2abc
   \]
   \[
   2abc \geq 0
   \]
   This is true since \( a, b, c \) are non-negative. Therefore, \( S \geq 2 \).

3. **Proof of upper bound:**
   We need to show:
   \[
   \frac{9}{4} \geq \frac{2(a+b+c)}{(a+b)(b+c)(c+a)}
   \]
   Simplifying:
   \[
   9(a+b)(b+c)(c+a) \geq 8(a+b+c)
   \]
   \[
   9((a+b+c)(ab+bc+ca) - abc) \geq 8(a+b+c)
   \]
   \[
   9(a+b+c) - 9abc \geq 8(a+b+c)
   \]
   \[
   a+b+c \geq 9abc
   \]
   Using AM-GM inequality:
   \[
   1 = ab + bc + ca \geq 3\sqrt[3]{a^2b^2c^2} \implies \frac{1}{3} \geq \sqrt[3]{a^2b^2c^2}
   \]
   \[
   a+b+c \geq 3\sqrt[3]{abc}
   \]
   It suffices to show:
   \[
   3\sqrt[3]{abc} \geq 9abc \implies \frac{1}{3} \geq \sqrt[3]{a^2b^2c^2}
   \]
   This holds from the established results. Therefore, \( S \leq \frac{9}{4} \).

4. **Find the minimum and maximum of \( f(S) = 3S - S^2 \):**
   We need to find the minimum and maximum of \( f(S) = 3S - S^2 \) in the interval \( 2 \leq S \leq \frac{9}{4} \).
   \[
   f(S) = 3S - S^2
   \]
   The function \( f(S) \) is a downward parabola with a maximum at \( S = \frac{3}{2} \). However, we are interested in the interval \( 2 \leq S \leq \frac{9}{4} \).

   Evaluating at the endpoints:
   \[
   f(2) = 3 \cdot 2 - 2^2 = 6 - 4 = 2
   \]
   \[
   f\left(\frac{9}{4}\right) = 3 \cdot \frac{9}{4} - \left(\frac{9}{4}\right)^2 = \frac{27}{4} - \frac{81}{16} = \frac{108}{16} - \frac{81}{16} = \frac{27}{16}
   \]

Therefore, the maximum value of the expression is \( 2 \) and the minimum value is \( \frac{27}{16} \).

The final answer is \( \boxed{2} \) and \( \boxed{\frac{27}{16}} \).