, 2013

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

Fix a positive integer $x$. Let for a positive integer $y$ greater than $x$,
$$b_y = \frac{a_x a_{x+1} + a_{x+1} a_{x+2} + \dots + a_{y-1} a_y}{a_x a_y}.$$
Then, we get
$$\begin{aligned} b_y + b_{y+2} &= \frac{1}{a_x} \left( \frac{a_x a_{x+1} + \dots + a_{y-1} a_y}{a_y} + \frac{a_x a_{x+1} + \dots + a_{y+1} a_{y+2}}{a_{y+2}} \right) \\ &= \frac{1}{a_x} \left( \frac{a_x a_{x+1} + \dots + a_{y-1} a_y}{a_y} + \frac{a_x a_{x+1} + \dots + a_y a_{y+1}}{a_{y+2}} + a_{y+1} \right) \\ &= \frac{1}{a_x} \left( \frac{a_x a_{x+1} + \dots + a_{y-1} a_y}{a_y} + \frac{a_y a_{y+1} + a_x a_{x+1}}{a_{y+2}} + \frac{a_x a_{x+1} + \dots + a_y a_{y+1}}{a_{y+2}} \right) \\ &= \frac{1}{a_x} \left( \frac{1}{a_y} + \frac{1}{a_{y+2}} \right) (a_x a_{x+1} + \dots + a_y a_{y+1}) \\ &= \left( \frac{a_{y+1}}{a_y} + \frac{a_y}{a_{y+2}} \right) \frac{a_x a_{x+1} + \dots + a_y a_{y+1}}{a_x a_y} \\ &= \left( \frac{a_{y+1}}{a_y} + \frac{a_y}{a_{y+2}} \right) b_{y+1}. \end{aligned}$$
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 $k = 0, 1, 2, \dots, m$, $\phi_k$ be the unique number lying in the interval $[0, 2\pi)$ for which $k\theta = 2\pi \cdot j + \phi_k$ is satisfied for some integer $j$. Since there are $m+1$ numbers $\phi_0, \phi_1, \dots, \phi_m$, 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 $0 \le p < q \le m$, $\phi_p$ and $\phi_q$ belong to the same interval. If $\phi_p = \phi_q$ holds, then we have $q\theta - p\theta = 2\pi \cdot (j - j')$ for some pair of integers $j$ and $j'$, so $(q-p)\theta$ is an integral multiple of $2\pi$ and we have
$$\frac{a_1 a_2 + \dots + a_{q-p} a_{p-q+1}}{a_1 a_{q-p+1}} = \frac{\sin(q-p)\theta}{\sin\theta} = 0,$$
$$\frac{a_1 a_2 + \dots + a_{q-p+1} a_{q-p+2}}{a_1 a_{q-p+2}} = \frac{\sin(q-p+1)\theta}{\sin\theta} = \frac{\sin\theta}{\sin\theta} = 1.$$
From the first of these equations we get $a_1 a_2 + \dots + a_{q-p} a_{q-p+1} = 0$, 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 $\phi_p \ne \phi_q$, which implies that $0 < |\phi_p - \phi_q| < \frac{2\pi}{m} < \pi - 2\alpha$. 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 $\phi_q - \phi_p$. 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
$$\frac{2}{\sqrt{4 - \left(\frac{a_2}{a_1} + \frac{a_2}{a_3}\right)^2}}.$$