First note that if $ \angle BIC_1 + \angle CIB_1 = 180^\circ $ than $ \angle C_1IB_1 + \angle CIB = 180^\circ $, so $\angle C_1IB_1 = 180^\circ -\angle CIB = \angle ABC/2 +\angle ACB/2$ so circumcircle of the triangle $BC_1I$ tangent to the circumcircle of triangle $B_1IC$ (Since $\angle C_1BI + \angle B_1CI =\angle ABC/2 +\angle ACB/2 = \angle C_1IB_1 $). Let they (circumcircles) meet $BC$ at points $K$ and $M$ respectively. Note that $ \angle KIM =\angle ABC/2 +\angle ACB/2$ from the tangency and so $\angle KIM = \angle C_1IB_1$. Also we see that $\angle IKC_1 = \angle IC_1K = IBK$ so $C_1I = IK$. Similarly $IM = IB_1$ so triangle $KIM$ equal to triangle $C_1IB_1$ so $C_1B_1 = KM$. By the second sparrow lemma $CK = AC - AB/2$ and $BM = AB - AC/2$ so we have $$ BC = BM + CK - KM = AB - AC/2 + AC - AB/2 - BC/2 $$or $$ AC+ AB = 3BC $$q.e.d.

Note that $$ \angle C_1IB_1=360^\circ-\angle CIB_1-\angle BIC_1-\angle BIC=180^\circ-\angle BIC=90^\circ-\frac{\angle BAC}{2}, $$which since $I$ lies on the bisector of $\angle BAC$ implies that $I$ is the $A$-excenter of $\triangle AB_1C_1$. It follows that $BC_1B_1C$ is circumscribed, thus $$ BC_1+CB_1=B_1C_1+BC\Leftrightarrow \frac{AB}{2}+\frac{AC}{2}=\frac{BC}{2}+BC\Leftrightarrow AB+AC=3BC. $$

This is very similar to CMIMC $2016$ Geometry Problem $7.$