China Girls' Mathematical Olympiad

By $x_{k+1}{x}_{k-1}-x_k^2=\frac{x_{k-1}^8-x_k^8}{(x_k{x}_{k-1}{)}^7}$, we have $$\frac{x_{k+1}}{x_k}-\frac{x_k}{x_{k-1}}=\frac{1}{x_k^8}-\frac{1}{x_{k-1}^8},$$ i.e. $$\frac{x_{k+1}}{x_k}-\frac{1}{x_k^8}=\frac{x_k}{x_{k-1}}-\frac{1}{x_{k-1}^8}=\cdots =\frac{x_2}{x_1}-\frac{1}{x_1^8}=\frac{7}{8}.$$ Hence, $x_{k+1}=\frac{7}{8}{x}_k+x_k^{-7}$, and when $x_1>0,x_k>0,k\ge 2$. By $x_{k+1}-x_k=x_k(x_k^{-8}-\frac{1}{8})$, we see that when $x_k^{-8}-\frac{1}{8}<0$, i.e. $x_k>8^{1/8}$, one has $x_{k+1}-x_k<0$, i.e. $x_{k+1}<x_k$, $k\ge 1$. And $x_{k+1}=\frac{7}{8}{x}_k+x_k^{-7}\ge 8\sqrt[8]{\frac{1}{8^7}}=8^{1/8}$, so when $x_k=8^{1/8}$, the equality holds. So if we take $a=8^{1/8}$, as soon as $x_k>8^{1/8}$ we have $$x_1>x_2>\cdots >x_n>\cdots .$$ When $x_1<8^{1/8}$, we have $x_2>x_1$ and $x_2>x_3>\cdots >x_n>\cdots$. So the constant that we are looking for is $a=8^{\frac{1}{8}}$.