Question 5
(ROM 1) Let $D$ be the interior of the circle $C$ and let $A\in C$. Show that the function $f:D\to\mathbb R$, $f(M)=\frac{|MA|}{|MM'|}$, where $M'=(AM\cap C$, is strictly convex; i.e., $f(P)\lt \frac{f(M_1)+f(M_2)}2$, $\forall M_1,M_2\in D$, $M_1\ne M_2$, where $P$ is the midpoint of the segment $M_1M_2$.