Upper bound on $\displaystyle\sum_{\text{cyc}}\dfrac{a}{a^3+b^2+c}$ Let $a,b,c$ be positive reals such that $a+b+c=3$.  Determine the largest possible value of $$\dfrac{a}{a^3+b^2+c}+\dfrac{b}{b^3+c^2+a}+\dfrac{c}{c^3+a^2+b}.$$
Experimenting for some values of $(a,b,c)$ I conjectured that the maximum value attained is $1$. But I am unable to prove this upper bound. Any hints or solutions are welcome.

Hint: 
Using Holder: $(a^3+b^2+c)(1+b+c)(1+1+c) \ge (a+b+c)^3$ to yield $$\frac{a}{a^3+b^2+c} \le \frac{a(1+b+c)(2+c)}{(a+b+c)^3}$$ and reduce the inequality to $5\sum ab + \sum a^2b + 3abc \le 21$ which is easy.
