Ireland_2017

Since $\cos x$ is decreasing on $[0,\pi /2]$ and $\sin x$ is increasing on this interval, by Chebyshev's inequality, $$\frac{\sin 2A+\sin 2B+\sin 2C}{\cos A+\cos B+\cos C}=\frac{2}{\cos A+\cos B+\cos C}\le \frac{2}{3}(\sin A+\sin B+\sin C).$$ But, just as in the solution of problem 25, $$\sin A+\sin B+\sin C=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$$ and, for all $x$, $y$, $z\in [0,\pi /2]$, $$\sqrt{\cos x\cos y\cos z}\le \cos \frac{x+y+z}{3},$$ with equality iff $x=y=z$. Hence, $$\frac{2}{3}(\sin A+\sin B+\sin C)\le \frac{8}{3}{\cos }^3\frac{A+B+C}{6}=\frac{8}{3}{\cos }^3\frac{\pi }{6}=\sqrt{3},$$ with equality iff $A=B=C=\pi /3$.