IMO Shortlist, 2006

We start with a preliminary observation. Let $T$ be a point inside the quadrilateral $ABCD$. Then:
$$\begin{align*} & \text{Circles } (BCT) \text{ and } (DAT) \text{ are tangent at } T \\ & \text{if and only if} \quad \angle ADT + \angle BCT = \angle ATB. \tag{1} \end{align*}$$
Indeed, if the two circles touch each other then their common tangent at $T$ intersects the segment $AB$ at a point $Z$, and so $\angle ADT = \angle ATZ$, $\angle BCT = \angle BTZ$, by the tangent-chord theorem. Thus $\angle ADT + \angle BCT = \angle ATZ + \angle BTZ = \angle ATB$.
And conversely, if $\angle ADT + \angle BCT = \angle ATB$ then one can draw from $T$ a ray $TZ$ with $Z$ on $AB$ so that $\angle ADT = \angle ATZ$, $\angle BCT = \angle BTZ$. The first of these equalities implies that $TZ$ is tangent to the circle $(DAT)$; by the second equality, $TZ$ is tangent to the circle $(BCT)$, so the two circles are tangent at $T$.

So the equivalence (1) is settled. It will be used later on. Now pass to the actual solution. Its key idea is to introduce the circumcircles of triangles $ABP$ and $CDP$ and to consider their second intersection $Q$ (assume for the moment that they indeed meet at two distinct points $P$ and $Q$).

Since the point $A$ lies outside the circle $(BCP)$, we have $\angle BCP + \angle BAP < 180^{\circ}$. Therefore the point $C$ lies outside the circle $(ABP)$. Analogously, $D$ also lies outside that circle. It follows that $P$ and $Q$ lie on the same $\operatorname{arc} CD$ of the circle $(BCP)$.

By symmetry, $P$ and $Q$ lie on the same arc $AB$ of the circle $(ABP)$. Thus the point $Q$ lies either inside the angle $BPC$ or inside the angle $APD$. Without loss of generality assume that $Q$ lies inside the angle $BPC$. Then
$$\begin{equation*} \angle AQD = \angle PQA + \angle PQD = \angle PBA + \angle PCD \leq 90^{\circ}. \tag{2} \end{equation*}$$
by the condition of the problem.

In the cyclic quadrilaterals $APQB$ and $DPQC$, the angles at vertices $A$ and $D$ are acute. So their angles at $Q$ are obtuse. This implies that $Q$ lies not only inside the angle $BPC$ but in fact inside the triangle $BPC$, hence also inside the quadrilateral $ABCD$.

Now an argument similar to that used in deriving (2) shows that
$$\begin{equation*} \angle BQC = \angle PAB + \angle PDC \leq 90^{\circ}. \tag{3} \end{equation*}$$
Moreover, since $\angle PCQ = \angle PDQ$, we get
$$\angle ADQ + \angle BCQ = \angle ADP + \angle PDQ + \angle BCP - \angle PCQ = \angle ADP + \angle BCP.$$
The last sum is equal to $\angle APB$, according to the observation (1) applied to $T = P$. And because $\angle APB = \angle AQB$, we obtain
$$\angle ADQ + \angle BCQ = \angle AQB.$$
Applying now (1) to $T = Q$ we conclude that the circles $(BCQ)$ and $(DAQ)$ are externally tangent at $Q$. (We have assumed $P \neq Q$; but if $P = Q$ then the last conclusion holds trivially.)

Finally consider the halfdiscs with diameters $BC$ and $DA$ constructed inwardly to the quadrilateral $ABCD$. They have centres at $M$ and $N$, the midpoints of $BC$ and $DA$ respectively. In view of (2) and (3), these two halfdiscs lie entirely inside the circles $(BQC)$ and $(AQD)$; and since these circles are tangent, the two halfdiscs cannot overlap. Hence $MN \geq \frac{1}{2} BC + \frac{1}{2} DA$.

On the other hand, since $\overrightarrow{MN} = \frac{1}{2}(\overrightarrow{BA} + \overrightarrow{CD})$, we have $MN \leq \frac{1}{2}(AB + CD)$. Thus indeed $AB + CD \geq BC + DA$, as claimed.