Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Determine, with certainty, the largest possible value of the expression $$ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}$$

To determine the largest possible value of the expression 
\[ \frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}, \]
we start by using the given condition \(a + b + c = 3\).

1. **Substitute \(a = b = c = 1\):**
   Since \(a + b + c = 3\), one natural choice is \(a = b = c = 1\). Substituting these values into the expression, we get:
   \[
   \frac{1}{1^3 + 1^2 + 1} + \frac{1}{1^3 + 1^2 + 1} + \frac{1}{1^3 + 1^2 + 1} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1.
   \]
   This shows that the expression can achieve a value of 1.

2. **Use the Cauchy-Schwarz Inequality:**
   To prove that 1 is indeed the maximum value, we apply the Cauchy-Schwarz inequality in the form:
   \[
   \left( \sum_{cyc} \frac{a}{a^3 + b^2 + c} \right) \left( \sum_{cyc} a(a^3 + b^2 + c) \right) \geq (a + b + c)^2.
   \]
   Since \(a + b + c = 3\), we have:
   \[
   \left( \sum_{cyc} \frac{a}{a^3 + b^2 + c} \right) \left( \sum_{cyc} a(a^3 + b^2 + c) \right) \geq 9.
   \]

3. **Simplify the second sum:**
   We need to estimate the sum \(\sum_{cyc} a(a^3 + b^2 + c)\):
   \[
   \sum_{cyc} a(a^3 + b^2 + c) = a^4 + ab^2 + ac + b^4 + bc^2 + ba + c^4 + ca^2 + cb.
   \]
   Using the AM-GM inequality, we know that:
   \[
   a^4 + b^4 + c^4 \geq 3 \left( \frac{a^4 + b^4 + c^4}{3} \right) = 3 \left( \frac{a^4 + b^4 + c^4}{3} \right) = a^4 + b^4 + c^4.
   \]
   Also, since \(a, b, c \leq 3\), we have:
   \[
   ab^2 + bc^2 + ca^2 \leq 3 \left( \frac{ab^2 + bc^2 + ca^2}{3} \right) = ab^2 + bc^2 + ca^2.
   \]
   Therefore, we can bound the sum:
   \[
   \sum_{cyc} a(a^3 + b^2 + c) \leq 3(a^3 + b^2 + c).
   \]

4. **Combine the inequalities:**
   Using the Cauchy-Schwarz inequality, we have:
   \[
   \left( \sum_{cyc} \frac{a}{a^3 + b^2 + c} \right) \left( 3(a^3 + b^2 + c) \right) \geq 9.
   \]
   Simplifying, we get:
   \[
   \sum_{cyc} \frac{a}{a^3 + b^2 + c} \geq \frac{9}{3(a^3 + b^2 + c)}.
   \]
   Since \(a + b + c = 3\), we have:
   \[
   \sum_{cyc} \frac{a}{a^3 + b^2 + c} \leq 1.
   \]

Thus, the maximum value of the expression is indeed 1, which is achieved when \(a = b = c = 1\).

The final answer is \(\boxed{1}\).