China Girls' Mathematical Olympiad

By $x_{k+1}x_{k-1} - x_k^2 = \frac{x_{k-1}^8 - x_k^8}{(x_kx_{k-1})^7}$, 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 $a=8^{\frac{1}{8}}$.