Ireland_2017

$$\begin{array}{rl}\sin 2x+\sin 2y+\sin 2z& =\sin 2x+\sin 2y-\sin 2(x+y)\\ & =\sin 2x(1-\cos 2y)+\sin 2y(1-\cos 2x)\\ & =2\sin 2x{\sin }^{2}y+2\sin 2y{\sin }^{2}x\\ & =4\sin x\sin y(\cos x\sin y+\cos y\sin x)\\ & =4\sin x\sin y\sin z.\end{array}$$ Similarly $$\begin{array}{rl}\sin x+\sin y+\sin z& =\sin x+\sin y+\sin (x+y)\\ & =\sin x(1+\cos y)+\sin y(1+\cos x)\\ & =2\sin x{\cos }^2\frac{y}{2}+2\sin y{\cos }^2\frac{x}{2}\\ & =4\cos \frac{x}{2}\cos \frac{y}{2}\left(\sin \frac{x}{2}\cos \frac{y}{2}+\sin \frac{y}{2}\cos \frac{x}{2}\right)\\ & =4\cos \frac{x}{2}\cos \frac{y}{2}\sin \frac{x+y}{2}\\ & =4\cos \frac{x}{2}\cos \frac{y}{2}\cos \frac{z}{2}.\end{array}$$ 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 $$\begin{array}{rl}2{\sin }^2\frac{a+b}{2}-2\sin a\sin b& =1-\cos (a+b)-2\sin a\sin b\\ & =1-\cos a\cos b-\sin a\sin b\\ & =1-\cos (a-b)\\ & \ge 0,\end{array}$$ 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.

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Alternative solution.

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 $△ABC$, $R$ 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 $$|OBC|=\frac{1}{2}{R}^2\sin 2x,|OCA|=\frac{1}{2}{R}^2\sin 2y,|OAB|=\frac{1}{2}{R}^2\sin 2z.$$ Using the well-known formula $|ABC|=rs$, these equations imply $$\begin{array}{rl}\sin 2x+\sin 2y+\sin 2z& =\frac{2}{R^2}(|OBC|+|OCA|+|OAB|)\\ & =\frac{2}{R^2}|ABC|=\frac{2rs}{R^2}.\end{array}$$ 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 $△ABC$. 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.