(Solution by Lenca Cuturela) Since $a + b + c = 1$, we have $a - a^2 = ab + ca$, so
$$\frac{bc + a + 1}{a^2 + 1} = 1 + \frac{bc + a - a^2}{a^2 + 1} = 1 + \frac{ab + bc + ca}{a^2 + 1},$$
and the similar relations. The inequality to be proven is equivalent with
$$\frac{1}{a^2 + 1} + \frac{1}{b^2 + 1} + \frac{1}{c^2 + 1} \le \frac{9}{10(ab + bc + ca)}. \quad (*)$$
Let $m = ab + bc + ca$; then $a^2 + b^2 + c^2 = 1 - 2m$ and, because $ab + bc + ca \le a^2 + b^2 + c^2$, we have $0 < m \le \frac{1}{3}$.
Since $\frac{1}{a^2 + 1} = 1 - \frac{a^2}{a^2 + 1}$, we can rewrite the inequality to be proven as
$$\frac{a^2}{a^2 + 1} + \frac{b^2}{b^2 + 1} + \frac{c^2}{c^2 + 1} \ge 3 - \frac{9}{10m}.$$
Using a variant of Cauchy-Schwarz inequality, we have
$$\frac{a^2}{a^2 + 1} + \frac{b^2}{b^2 + 1} + \frac{c^2}{c^2 + 1} \ge \frac{(a+b+c)^2}{a^2 + b^2 + c^2 + 3} = \frac{1}{4-2m}.$$
It is enough to prove that $\frac{1}{4-2m} \ge 3 - \frac{9}{10m}$, that is $(3m-1)(5m-9) \ge 0$, which is true, because $m \in (0, \frac{1}{3}]$. The equality holds when $a = b = c$ and $m = \frac{1}{3}$, so $a = b = c = \frac{1}{3}$.