By using the condition $a+b+c=1$ and the inequality between arithmetic and geometric means, we have that
$$\frac{a}{a+b^2} = \frac{a}{a(a+b+c)+b^2} = \frac{a}{a^2+b^2+ab+ac} \le \frac{a}{2ab+ab+ac} = \frac{1}{3b+c}.$$
By applying the inequality between harmonic and arithmetic means, it follows that
$$\frac{4}{\frac{3}{b} + \frac{1}{c}} = \frac{4}{\frac{1}{b} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c}} \le \frac{b+b+b+c}{4} = \frac{3b+c}{4},$$
i.e.
$$\frac{a}{a+b^2} \le \frac{1}{3b+c} \le \frac{1}{16} \left( \frac{3}{b} + \frac{1}{c} \right).$$
Analogously,
$$\frac{b}{b+c^2} \le \frac{1}{16} \left( \frac{3}{c} + \frac{1}{a} \right) \quad \text{and} \quad \frac{c}{c+a^2} \le \frac{1}{16} \left( \frac{3}{a} + \frac{1}{b} \right).$$
Finally, by adding the inequalities above, we have that
$$\frac{a}{a+b^2} + \frac{b}{b+c^2} + \frac{c}{c+a^2} \le \frac{1}{16} \left( \frac{4}{a} + \frac{4}{b} + \frac{4}{c} \right) = \frac{1}{4} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right).$$