Irish Mathematical Olympiad

$$\begin{align*} \sin 2x + \sin 2y + \sin 2z &= \sin 2x + \sin 2y - \sin 2(x + y) \\ &= \sin 2x(1 - \cos 2y) + \sin 2y(1 - \cos 2x) \\ &= 2 \sin 2x \sin^2 y + 2 \sin 2y \sin^2 x \\ &= 4 \sin x \sin y (\cos x \sin y + \cos y \sin x) \\ &= 4 \sin x \sin y \sin z. \end{align*}$$
Similarly
$$\begin{align*} \sin x + \sin y + \sin z &= \sin x + \sin y + \sin(x + y) \\ &= \sin x(1 + \cos y) + \sin y(1 + \cos x) \\ &= 2 \sin x \cos^2 \frac{y}{2} + 2 \sin y \cos^2 \frac{x}{2} \\ &= 4 \cos \frac{x}{2} \cos \frac{y}{2} \left( \sin \frac{x}{2} \cos \frac{y}{2} + \sin \frac{y}{2} \cos \frac{x}{2} \right) \\ &= 4 \cos \frac{x}{2} \cos \frac{y}{2} \sin \frac{x+y}{2} \\ &= 4 \cos \frac{x}{2} \cos \frac{y}{2} \cos \frac{z}{2}. \end{align*}$$
Hence
$$\frac{\sin 2x + \sin 2y + \sin 2z}{\sin x + \sin y + \sin z} = 8 \sin \frac{x}{2} \sin \frac{y}{2} \sin \frac{z}{2}.$$
But, if $0 < a, b < \pi$, then
$$\sqrt{\sin a \sin b} \le \sin \frac{a+b}{2},$$
since
$$\begin{align*} 2 \sin^2 \frac{a+b}{2} - 2 \sin a \sin b &= 1 - \cos(a+b) - 2 \sin a \sin b \\ &= 1 - \cos a \cos b - \sin a \sin b \\ &= 1 - \cos(a-b) \\ &\ge 0, \end{align*}$$
with equality iff $a = b$. Using Jensen's Inequality for the function $f(x) = \ln(\sin x)$ it follows that if $a, b, c \in [0, \pi]$, then
$$\sqrt[3]{\sin a \sin b \sin c} \le \sin \frac{a+b+c}{3}.$$
Moreover, the inequality is strict unless $a = b = c$.
It follows that
$$\sqrt[3]{\sin \frac{x}{2} \sin \frac{y}{2} \sin \frac{z}{2}} \le \sin \frac{x+y+z}{6} = \sin \frac{\pi}{6} = \frac{1}{2}$$
whence
$$8 \sin \frac{x}{2} \sin \frac{y}{2} \sin \frac{z}{2} \le 1,$$
with equality iff $x = y = z = \pi/3$. The result follows.

Let $ABC$ be a triangle with angles of size $x$, $y$ and $z$ at $A$, $B$ and $C$, respectively. We follow standard notation and let $O$ be the circumcentre of $\triangle ABC$, $R$ the circumradius, $r$ the inradius, $a, b, c$ the side lengths and $s$ the semi-perimeter.
By $|OBC|$ we denote the area of triangle $OBC$ etc. Because $|OA| = |OB| = |OC| = R$ and $\angle BOC = 2\angle BAC$ (central angle) etc., we have
$$|OBC| = \frac{1}{2} R^2 \sin 2x, \quad |OCA| = \frac{1}{2} R^2 \sin 2y, \quad |OAB| = \frac{1}{2} R^2 \sin 2z.$$
Using the well-known formula $|ABC| = r s$, these equations imply
$$\begin{align*} \sin 2x + \sin 2y + \sin 2z &= \frac{2}{R^2} (|OBC| + |OCA| + |OAB|) \\ &= \frac{2}{R^2} |ABC| = \frac{2 r s}{R^2}. \end{align*}$$
On the other hand, from the extended Sine-Rule
$$\frac{\sin x}{a} = \frac{\sin y}{b} = \frac{\sin z}{c} = \frac{1}{2R}$$
we obtain
$$\sin x + \sin y + \sin z = \frac{1}{2R}(a + b + c) = \frac{s}{R}.$$
Therefore,
$$\frac{\sin 2x + \sin 2y + \sin 2z}{\sin x + \sin y + \sin z} = \frac{2r}{R}$$
and the desired inequality is equivalent to Euler's inequality $R \ge 2r$, which is a consequence of Euler's Theorem $|OI|^2 = R(R-2r)$, where $I$ is the incentre of $\triangle ABC$.
The case of equality, $R = 2r$, therefore occurs exactly when $O = I$ and this is easily seen to be the case iff the triangle $ABC$ is equilateral.