, 2006

Let $\angle A = \alpha$, $\angle B = \beta$, $\angle C = \gamma$. Since $\angle PBA + \angle PCA + \angle PBC + \angle PCB = \beta + \gamma$, the condition from the problem statement is equivalent to $\angle PBC + \angle PCB = (\beta + \gamma)/2$, i.e. $\angle BPC = 90^\circ + \alpha/2$.

On the other hand, $\angle BIC = 180^\circ - (\beta + \gamma)/2 = 90^\circ + \alpha/2$. Hence $\angle BPC = \angle BIC$, and since $P$ and $I$ are on the same side of $BC$, the points $B$, $C$, $I$ and $P$ are concyclic. In other words, $P$ lies on the circumcircle $\omega$ of triangle $BCI$.

Let $\Omega$ be the circumcircle of triangle $ABC$. It is a well-known fact that the centre of $\omega$ is the midpoint $M$ of the arc $BC$ of $\Omega$. This is also the point where the angle bisector $AI$ intersects $\Omega$.

From triangle $APM$ we have
$$AP + PM \geq AM = AI + IM = AI + PM$$
Therefore $AP \geq AI$. Equality holds if and only if $P$ lies on the line segment $AI$, which occurs if and only if $P = I$.