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 $A = B = C = \pi/3$.