Let
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
be the interior of the circle
![C](/media/m/5/a/b/5ab88f3f735b691e133767fe7ea0483c.png)
and let
![A \in C](/media/m/6/5/1/651a0fc16e094a8b4e0d2599c13ac180.png)
. Show that the function
![f : D \to \mathbb R, f(M)=\frac{|MA|}{|MM'|}](/media/m/3/8/e/38efddab0419e47d0814bdcfd628a397.png)
where
![M' = AM \cap C](/media/m/2/2/3/22373f0ae39412f77a6720dc4ebefe2d.png)
, is strictly convex; i.e.,
![f(P) <\frac{f(M_1)+f(M_2)}{2}, \forall M_1,M_2 \in D, M_1 \neq M_2](/media/m/2/1/f/21fff093a71b6c0de4af77647c6253db.png)
where
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
is the midpoint of the segment
%V0
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., $f(P) <\frac{f(M_1)+f(M_2)}{2}, \forall M_1,M_2 \in D, M_1 \neq M_2$ where $P$ is the midpoint of the segment $M_1M_2.$