SAUDI ARABIAN MATHEMATICAL COMPETITIONS

From $O,A_1,A_{2016}$ are collinear, there exists some positive number $k$ such that $$\frac{x_1}{y_1}=\frac{x_{2016}}{y_{2016}}=k>0.$$ We shall prove that for all $i=\overline{2,2015}$, $\frac{x_i}{y_i}>k$. Indeed, Notice that $$x_{n+1}=\sqrt{\frac{x_n^2+x_{n+2}^2}{2}}\iff x_{n+1}^2=\frac{x_n^2+x_{n+2}^2}{2},$$ which implies that $(x_n^2)$ forms an arithmetic sequence. Hence, for all $i=\overline{2,2015}$, let $${\alpha }_i=\frac{2016-i}{2015},{\beta }_i=\frac{i-1}{2015}$$ then ${\alpha }_i+{\beta }_i=1$, $2\le i\le 2015$ and $$x_i^2={\alpha }_i{x}_1^2+{\beta }_i{x}_{2016}^2\iff x_i=\sqrt{{\alpha }_i{x}_1^2+{\beta }_i{x}_{2016}^2}.$$ Similarly, we also have $$y_i={\left({\alpha }_i\sqrt{y_1}+{\beta }_i\sqrt{y_{2016}}\right)}^2\text{ and }k{y}_i={\left({\alpha }_i\sqrt{x_1}+{\beta }_i\sqrt{x_{2016}}\right)}^2.$$ We need to prove $$\sqrt{{\alpha }_i{x}_1^2+{\beta }_i{x}_{2016}^2}>{\left({\alpha }_i\sqrt{x_1}+{\beta }_i\sqrt{x_{2016}}\right)}^2$$ for all $2\le i\le 2015$ $(\ast )$. By applying the Cauchy-Schwarz inequality, we have $$\begin{array}{rl}({\alpha }_i+{\beta }_i)({\alpha }_i{x}_1^2+{\beta }_i{x}_{2016}^2)& \ge ({\alpha }_i{x}_1+{\beta }_i{x}_{2016}{)}^2\\ & ={\left(({\alpha }_i+{\beta }_i)({\alpha }_i{x}_1+{\beta }_i{x}_{2016})\right)}^2\\ & \ge {\left({\alpha }_i\sqrt{x_1}+{\beta }_i\sqrt{x_{2016}}\right)}^4\end{array}$$ So the inequality $(\ast )$ is true for each $2\le i\le 2015$. Since all points are distinct, then $A_1\ne A_{2016}\iff x_1\ne x_{2016}$, so equality does not occur in $(\ast )$. So all the points $A_2,A_3,\dots ,A_{2015}$ belong to the same side with respect to $d$.