Find the maximal value of the following expression, if $a,b,c$ are nonnegative and $a+b+c=1$.
\[ \frac{1}{a^2 -4a+9} + \frac {1}{b^2 -4b+9} + \frac{1}{c^2 -4c+9} \]
1. **Claim the Maximum Value**:
   We claim that the maximum value of the given expression is \(\boxed{\frac{7}{18}}\). This is achieved when \((a, b, c) = (1, 0, 0)\) and its permutations.

2. **Expression to Prove**:
   We need to prove that:
   \[
   \frac{1}{a^2 - 4a + 9} \leq \frac{a + 2}{18}
   \]
   for \(a \in [0, 1]\).

3. **Multiply Both Sides**:
   Multiply both sides by \(18(a^2 - 4a + 9)\):
   \[
   18 \leq (a + 2)(a^2 - 4a + 9)
   \]

4. **Expand and Simplify**:
   Expand the right-hand side:
   \[
   18 \leq a^3 - 4a^2 + 9a + 2a^2 - 8a + 18
   \]
   Simplify the expression:
   \[
   18 \leq a^3 - 2a^2 + a + 18
   \]

5. **Subtract 18 from Both Sides**:
   \[
   0 \leq a^3 - 2a^2 + a
   \]

6. **Factor the Polynomial**:
   Factor the right-hand side:
   \[
   0 \leq a(a^2 - 2a + 1)
   \]
   \[
   0 \leq a(a - 1)^2
   \]

7. **Analyze the Inequality**:
   Since \(a \geq 0\) and \((a - 1)^2 \geq 0\), the product \(a(a - 1)^2 \geq 0\) is always true for \(a \in [0, 1]\).

8. **Summing Up**:
   Since \(\frac{1}{a^2 - 4a + 9} \leq \frac{a + 2}{18}\) holds for \(a \in [0, 1]\), we can sum the inequalities for \(a, b, c\):
   \[
   \frac{1}{a^2 - 4a + 9} + \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9} \leq \frac{a + 2}{18} + \frac{b + 2}{18} + \frac{c + 2}{18}
   \]
   Since \(a + b + c = 1\), we have:
   \[
   \frac{a + 2}{18} + \frac{b + 2}{18} + \frac{c + 2}{18} = \frac{(a + b + c) + 6}{18} = \frac{1 + 6}{18} = \frac{7}{18}
   \]

Thus, the maximum value of the given expression is \(\boxed{\frac{7}{18}}\).