Inequality for sides of a triangles I recently encountered this geometric problem (I could not find a better title). 

Let $ABC$ a triangle and $P$ a point on the segment $AB$. Let $Q$ be
  the intersection of the line $CP$ with the circumscribed circle of the
  triangle different from $C$.   Show the inequality
  $$\frac{\overline{PQ}}{\overline{CQ}} \le \Big(\frac
 {\overline{AB}}{\overline{AC}+\overline{CB}}\Big)^2 .$$   Further,
  show that equality holds if and only if $CP$ is the angle bisector of
  the angle $\angle ACB$.

As in most problems of this kind, finding the solution is a matter drawing the right lines. I already come up with a solution. I wanted to share the problem as there are probably many different ways to solve it.
To be clear: I don't need help with it, I just thought maybe there are some people here who like to do this kind of problems and want to try themselves on it.

 A: First, let $CPQ$ bisect $\angle ACB$, and join $AQ$, $BQ$. Then $AQ=BQ$, and by Ptolemy's Theorem $$AB\times CQ=AQ\times(CA+CB)$$Dividing both sides by $AB\times AQ$, inverting and squaring$$\frac{AQ^2}{CQ^2}=\frac{AB^2}{(AC+CB)^2}$$Thus we must show that $$\frac{PQ}{CQ}=\frac{AQ^2}{CQ^2}=\frac{AB^2}{(AC+CB)^2}$$or$$PQ\times CQ=AQ^2$$But since $\angle ACB$ is bisected, $\angle ACQ=\angle QAP$, making triangles $ACQ$ and $QAP$ similar.
Hence$$\frac{PQ}{AQ}=\frac{AQ}{CQ}$$or$$PQ\times CQ=AQ^2$$
Thus the ratios are equal when $\angle ACB$ is bisected.
Now let $R$ be any other point on $AB$, and draw $CR$ thru to $S$ on the circumference, and join $QS$. 
Then since arcs $AQ$, $BQ$ are equal, the tangent at $Q$ is parallel to $AB$, and therefore chord $QS$ is inclined toward $PR$, making$$\frac{SR}{RC}<\frac{QP}{PC}$$Hence$$\frac{SR}{SR+RC}<\frac{QP}{QP+PC}$$or$$\frac{SR}{SC}<\frac{QP}{QC}$$
And since $\frac{AB^2}{(AC+CB)^2}$ is fixed$$\frac{SR}{SC}<\frac{AB^2}{(AC+CB)^2}$$
A: Here is another proof which doesn't use Ptolemy's. 
The idea is to replace both sides of the inequality with the area of some object and then use congruent triangles in order to show that the areas agree if $Q$ is on the angle bisector of $\angle ACB$.
We start noting that in the quadrilateral $\Box CAQB$, the point $P$ is the intersection of its diagonals and one diagonal is a edge of $\triangle AQB$.
Let $h_1$ the height of the triangle $\triangle ABC$ and $h_2$ the height of the triangle $\triangle AQB$, both with respect to the base $AB$. Then the areas denoted by $\mathcal A$ satisfy
$$\frac{\mathcal A(\Box CAQB)}{\mathcal A(\triangle AQB)}=\frac{\mathcal A(\triangle AQB)+\mathcal A(\triangle ABC)}{\mathcal A(\triangle AQB)} = 1+\frac{h_1}{h_2}=1+\frac{\overline{CP}}{\overline{PQ}}= \frac{\overline{CQ}}{\overline{PQ}},  $$
where in the second to last equality we used the intercept theorem.
This further shows that (for fixed $\triangle ABC$) the ratio $\frac{\overline{PQ}}{\overline{CQ}}$ is maximal iff ${\mathcal A(\triangle AQB)}$ is maximal i.e. $h_2$ is maximal. But this is if $Q$ is on the line segment bisector of $AB$. (Then the parallel of $AB$ through $Q$ is tangential to the circumscribed circle). As line segment bisectors and angle bisectors intersect on the circumscribed circle, this shows that $\frac{\overline{PQ}}{\overline{CQ}}$ is maximal iff $Q$ is on the angle bisector.
In order to simplify the right-hand side, consider the scaling of the triangle $\triangle AQB$ with center $C$ by the factor $k=\frac{\overline{AB}}{\overline{AC}+\overline{CB}}$. We get a quadrilateral $\Box CDEF$ with the property that 
$$\frac{\mathcal A(\Box CDEF)}{\mathcal A(\Box CAQB)}=k^2=\Big(\frac{\overline{AB}}{\overline{AC}+\overline{CB}}\Big)^2.$$
Thus we have reduced the problem to showing that 
$$\frac{\mathcal A(\triangle AQB)}{\mathcal A(\Box CAQB)}\le \frac{\mathcal A(\Box CDEF)}{\mathcal A(\Box CAQB)}.$$
As we already proved the criterium for the left-hand side to be maximal, it suffices to show that 
$${\mathcal A(\triangle AQB)} = {\mathcal A(\Box CDEF)}, $$
if $Q$ is on the angle bisector of $\angle ACB$.

So let $CQ$ the angle bisector of $\angle ACB$. We show that ${\mathcal A(\triangle AQB)} = {\mathcal A(\Box CDEF)}$, by showing that the triangles $\triangle AQP$ and $\triangle DEC$, resp $\triangle PQB$ and $\triangle CEF$ are congruent. As the cases are analogous, we only need to show the first congruency.
By construction, the segments $AQ$ and $DE$ are parallel. Thus we have $\angle PQA=\angle CED.$
Further the angles $\angle DCE$ and $\angle QAP$ are equal because they are angles over the common chord $QB$.
So it is left to show that two edges of the triangles have the same length. But the angle bisector of $\angle ACB$ divides $AB$ in the ratio of the adjacent sides $AC$ and $BC$. So $$\frac{\overline{AP}}{\overline{PB}}= \frac{\overline{AC}}{\overline{BC}},$$ which implies $$\frac{\overline{DC}}{\overline{AC}} =k= \frac{\overline{AB}}{\overline{AC}+\overline{CB}} = \frac{\overline{AB}}{\overline{AP}} \cdot\frac{\overline{AP}}{\overline{AC}+\overline{CB}} = \frac{\overline{AC}+\overline{CB}}{\overline{AC}} \cdot\frac{\overline{AP}}{\overline{AC}+\overline{CB}}= \frac{\overline{AP}}{\overline{AC}}, $$
so $\overline{AP}=\overline{DC}$.
This finishes the proof.
