$r=1$, $R=OC/OB$. Unit ball is $DEBD'$ and symmetric part, $M$ is $DCD'$ and symmetric part, $x=B$, $y=G$, $\pi(x)=C$, $\pi(y)=H$, ray $OGH$ is very close to $OBC$.
We have $$\|C-H\|:\|B-G\|=\frac{CH}{OI}:\frac{BG}{OD}=\frac{CH\cdot BE}{BG\cdot CE}\cdot \frac{CE}{OI}\cdot \frac{OD}{BE}=\frac{OH}{OG}\cdot \frac{CE}{OI}\cdot\frac{OC}{CB}$$ (Menelaus theorem for triangle $EHG$ and line $CBO$ is used.) $OH/OG$ tends to $OC/OB$ when the ray $OGH$ tends to $OCB$. Next, $CE/OI=(CE/ED)\cdot (ED/OI)=(CB/BO)\cdot (1+OB/OC)$. Totally we get in the limit $(OC/OB)^2+(OC/OB)=R^2+R$.
We have used relation $ED/OI=1+OB/OC$, which follows, for example, from parallelogram $EBD'M$: $ED/OI=DM/OD'=(2DO-EB)/DO=2-EB/DO=2-CB/OC=1+OB/OC$.
