Baltic Way shortlist

Let the area of the triangle $△\Delta$ be $S$. Among the triangles with fixed circumradius, the one with largest perimeter is equilateral (as can be easily seen from Jensen's inequality). Hence $$S=\frac{a+b+c}{2}\cdot r\le \frac{3\sqrt{3}}{2}Rr.$$ By Euler's inequality, $R\ge 2r$. Thus $$R^2+r^2=\frac{3}{4}{R}^2+\left(\frac{1}{4}{R}^2+r^2\right)\ge \frac{3}{2}Rr+Rr=\frac{5}{2}Rr.$$ Hence $$\pi (R+r{)}^2\ge \pi \cdot \frac{9}{2}Rr>5\cdot \frac{3\sqrt{3}}{2}Rr\ge 5S.$$