$\definecolor{A}{RGB}{162,156,165}\color{A}\fbox{Solution}$ Let $A,B,C,D,E$ be the consecutive vertices of pentagon lying on the circle and $O$ be the circumcircle. Denote $\alpha_1=\angle AOB,\ \alpha_2=\angle BOC,\ ...,\alpha_5=\angle EOA$ and $\alpha_1=\alpha_6$, all the angles belong to interval $(0^\circ,180^\circ)$. Since $O$ can lie outside of convex pentagon $ABCDE$ we have $$[ABCDE]\le[AOB]+[BOC]+...+[EOA]=\frac12R^2\sum_{i=1}^5\sin\alpha_i.$$From the inscribed angle theorem and sine theorem we have that sum of diagonals is equal to $$S=2R\sum_{i=1}^5\sin\frac{\alpha_i+\alpha_{i+1}}{2}.$$Observe $$\forall_{i\in\lbrace 1,2,3,4,5\rbrace} \left[\frac{\alpha_{i}+\alpha_{i+1}}{2}\in(0^\circ,180^\circ)\implies\frac{\sin\alpha_i+\sin\alpha_{i+1}}{2}=\sin\frac{\alpha_{i}+\alpha_{i+1}}{2}\cos\frac{\alpha_{i}-\alpha_{i+1}}{2}\le \sin\frac{\alpha_{i}+\alpha_{i+1}}{2}\right].$$Therefore $$[ABCDE]\le\frac12R^2\sum_{i=1}^5\sin\alpha_i=\frac12R^2\sum_{i=1}^5\frac{\sin\alpha_i+\sin\alpha_{i+1}}{2}\le\frac12R^2\sum_{i=1}^5\sin\frac{\alpha_{i}+\alpha_{i+1}}{2}=\frac14R\cdot S.\blacksquare$$#1865

The official solution uses a different argument.