Japan 2013 Initial Round

$$\boxed{\frac{2}{\sqrt{4-{\left(\frac{a_2}{a_1}+\frac{a_2}{a_3}\right)}^2}}}$$

Fix a positive integer $x$. Let for a positive integer $y$ greater than $x$, $$b_y=\frac{a_x{a}_{x+1}+a_{x+1}{a}_{x+2}+\cdots +a_{y-1}{a}_y}{a_x{a}_y}.$$ Then, we get $$\begin{array}{rl}{b}_y+b_{y+2}& =\frac{1}{a_x}\left(\frac{a_x{a}_{x+1}+\cdots +a_{y-1}{a}_y}{a_y}+\frac{a_x{a}_{x+1}+\cdots +a_{y+1}{a}_{y+2}}{a_{y+2}}\right)\\ & =\frac{1}{a_x}\left(\frac{a_x{a}_{x+1}+\cdots +a_{y-1}{a}_y}{a_y}+\frac{a_x{a}_{x+1}+\cdots +a_y{a}_{y+1}}{a_{y+2}}+a_{y+1}\right)\\ & =\frac{1}{a_x}\left(\frac{a_x{a}_{x+1}+\cdots +a_{y-1}{a}_y}{a_y}+\frac{a_y{a}_{y+1}+a_x{a}_{x+1}}{a_{y+2}}+\frac{a_x{a}_{x+1}+\cdots +a_y{a}_{y+1}}{a_{y+2}}\right)\\ & =\frac{1}{a_x}\left(\frac{1}{a_y}+\frac{1}{a_{y+2}}\right)(a_x{a}_{x+1}+\cdots +a_y{a}_{y+1})\\ & =\left(\frac{a_{y+1}}{a_y}+\frac{a_y}{a_{y+2}}\right)\frac{a_x{a}_{x+1}+\cdots +a_y{a}_{y+1}}{a_x{a}_y}\\ & =\left(\frac{a_{y+1}}{a_y}+\frac{a_y}{a_{y+2}}\right)b_{y+1}.\end{array}$$ By assumption, $\frac{a_{y+1}}{a_y}+\frac{a_{y+1}}{a_{y+2}}=\left(\frac{a_2}{a_1}+\frac{a_2}{a_3}\right)$ is a constant bigger than 0 and less than 2, so we can write this constant as $2\cos \theta$ by using an angle $\theta$ satisfying $0<\theta <\frac{\pi }{2}$. The identity we obtained above can be represented as $$b_y+b_{y+2}=2\cos \theta \cdot b_{y+1}.$$ Also, by letting $b_x=0$ and $y=x$, we get $b_{x+1}=\frac{a_x{a}_{x+1}}{a_x{a}_{x+1}}=1$, and $b_{x+2}=2\cos \theta$. Using the identity $$\sin \theta +\sin (n+2)\theta =2\cos \theta \sin (n+1)\theta ,$$ we can prove by mathematical induction that $b_{x+n}=\frac{\sin n\theta }{\sin \theta }$ holds for any non-negative integer $n$. Consequently, to obtain the desired answer to the problem it suffices to find the smallest positive number $c$ which satisfies the inequality $$\frac{\sin n\theta }{\sin \theta }\le c$$ for any $n\ge 1$. Next, we show that if a positive constant $c$ satisfies $c>\frac{\sin n\theta }{\sin \theta }$ for any $n\ge 1$, then for any $\alpha$ satisfying $0<\alpha <\frac{\pi }{2}$, we must have $c>\frac{\sin \alpha }{\sin \theta }$. To show this, let us choose a positive integer $m$ large enough so that $\frac{2\pi }{m}<\pi -2\alpha$. Let for $k=0,1,2,\dots ,m$, ${\varphi }_k$ be the unique number lying in the interval $[0,2\pi )$ for which $k\theta =2\pi \cdot j+{\varphi }_k$ is satisfied for some integer $j$. Since there are $m+1$ numbers ${\varphi }_0,{\varphi }_1,\dots ,{\varphi }_m$, by the pigeon-hole principle, there exists at least one interval among $m$ disjoint intervals $[\frac{2\pi (i-1)}{m},\frac{2\pi i}{m})$ ($i=1,2,\dots ,m$), which contains two or more of these $m+1$ numbers. Suppose for some $p,q$ with $0\le p<q\le m$, ${\varphi }_p$ and ${\varphi }_q$ belong to the same interval. If ${\varphi }_p={\varphi }_q$ holds, then we have $q\theta -p\theta =2\pi \cdot (j-j^{\prime })$ for some pair of integers $j$ and $j^{\prime }$, so $(q-p)\theta$ is an integral multiple of $2\pi$ and we have $$\frac{a_1{a}_2+\cdots +a_{q-p}{a}_{p-q+1}}{a_1{a}_{q-p+1}}=\frac{\sin (q-p)\theta }{\sin \theta }=0,$$ $$\frac{a_1{a}_2+\cdots +a_{q-p+1}{a}_{q-p+2}}{a_1{a}_{q-p+2}}=\frac{\sin (q-p+1)\theta }{\sin \theta }=\frac{\sin \theta }{\sin \theta }=1.$$ From the first of these equations we get $a_1{a}_2+\cdots +a_{q-p}{a}_{q-p+1}=0$, and substituting this into the second equation, we get $a_{q-p+1}=a_1$. But this contradicts the assumption that all the $a_j$'s are distinct. Therefore, we have ${\varphi }_p\ne {\varphi }_q$, which implies that $0<|{\varphi }_p-{\varphi }_q|<\frac{2\pi }{m}<\pi -2\alpha$. Now if for any $k\ge 1$, we write $k(q-p)\theta =2\pi s+{\eta }_k$, where $0\le {\eta }_k<2\pi$ and $s$ is some integer, then as $k$ increases by 1, ${\eta }_k$ changes by the amount ${\varphi }_q-{\varphi }_p$. Therefore, there exists a positive integer $k$ so that ${\eta }_k$ belongs to the interval $(\alpha ,\pi -\alpha )$. For this $k$ we have $\sin k(q-p)\theta =\sin (2\pi s+{\eta }_k)=\sin {\eta }_k>\sin \alpha$, and therefore, we get $$c\ge \frac{\sin k(q-p)\theta }{\sin \theta }>\frac{\sin \alpha }{\sin \theta }.$$ Thus we conclude that $c>\frac{\sin \alpha }{\sin \theta }$ holds for any $\alpha$ satisfying $0<\alpha <\frac{\pi }{2}$. From this we can conclude also that $c\ge \frac{1}{\sin \theta }$ is satisfied. If we let $c_0=\frac{1}{\sin \theta }$, then $\frac{\sin n\theta }{\sin \theta }\le c_0$ obviously holds. From these considerations we conclude that the smallest constant $c$ which satisfies the requirement of the problem equals $c_0=\frac{1}{\sin \theta }=\frac{1}{\sqrt{1-{\cos }^2\theta }}$, and since $2\cos \theta =(\frac{a_2}{a_1}+\frac{a_2}{a_3})$, the desired answer for the problem is given by $$\frac{2}{\sqrt{4-{\left(\frac{a_2}{a_1}+\frac{a_2}{a_3}\right)}^2}}.$$