Maximum

I am pretty sure this is not the best solution available, so keep trying!

Observe that, $\large {\frac {a}{a^3+b^2+c}+\frac {b}{b^3+c^2+a}+\frac {c}{c^3+a^2+b}= \frac {1}{a^2+\frac {b^2}{a}+\frac {c}{a}}+\frac {1}{b^2+\frac{c^2}{b}+\frac {a}{b}}+\frac {1}{c^2+\frac {a^2}{c}+\frac {b}{c}}}$ .

By the Cauchy-Schwarz Inequality ,

$(a^2+\frac{b^2}{a}+\frac{c}{a})(1+a+ac) \geq (a+b+c)^2=9 \implies a^2+\frac{b^2}{a}+\frac{c}{a} \geq \frac{9}{1+a+ac} \implies \large{\frac{1}{a^2+\frac{b^2}{a}+\frac{c}{a}} \leq \frac{1+a+ac}{9}}$ .

Similarly, $\large{\frac {1}{b^2+\frac{c^2}{b}+\frac {a}{b}} \leq \frac{1+b+ba}{9}}$ and $\large{\frac {1}{c^2+\frac {a^2}{c}+\frac {b}{c}} \leq \frac{1+c+cb}{9}}.$

Adding these resuts we get, $\large {\frac {a}{a^3+b^2+c}+\frac {b}{b^3+c^2+a}+\frac {c}{c^3+a^2+b}= \frac {1}{a^2+\frac {b^2}{a}+\frac {c}{a}}+\frac {1}{b^2+\frac{c^2}{b}+\frac {a}{b}}+\frac {1}{c^2+\frac {a^2}{c}+\frac {b}{c}} \leq \frac{1+1+1+a+b+c+ab+bc+ca}{9}=\frac{6+ab+bc+ca}{9}.}$ .

Now the claim is that $ab+bc+ca \leq 3.$ If we substitute this in the above inequality we get $\large {\frac {a}{a^3+b^2+c}+\frac {b}{b^3+c^2+a}+\frac {c}{c^3+a^2+b}} \leq \boxed{1}$

Proof of the claim that $ab+bc+ca \leq 3$ is left to the readers :) Hint: Use $2(ab+bc+ca)=(a+b+c)^2-a^2-b^2-c^2$