= Solution
The arithmetic--geometric mean inequality gives $x_n\geq1$ from $n=2$ onward. For $x>1$,
$$
1<\frac12(x+x^{-1})<x,
$$
so the tail decreases to a <limit> $L\geq1$. Passing to the recurrence gives $2L=L+L^{-1}$, hence $L=1$.
For the subadditive <sequence>, $0\leq x_n/n\leq x_1$, so it is bounded. The <Fekete lemma> can be proved directly here as follows. Let $\alpha=\inf_kx_k/k$. Fix $k$ and write $n=qk+r$, $0\leq r<k$. Then
$$
\frac{x_n}{n}\leq\frac{qx_k+x_r}{qk+r}\longrightarrow\frac{x_k}{k}.
$$
\b[Thus $\limsup x_n/n\leq\alpha$, while the definition gives $\liminf x_n/n\geq\alpha$. Hence $x_n/n\to\alpha$.]
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