 Solution

The arithmetic--geometric mean inequality gives $x_n\ge 1$ from $n=2$ onward. For $x>1$,so the tail decreases to a limit $L\ge 1$. Passing to the recurrence gives $2L=L+L^{-1}$, hence $L=1$.

For the subadditive sequence, $0\le x_n/n\le x_1$, so it is bounded. The Fekete lemma can be proved directly here as follows. Let $\alpha ={\inf }_k{x}_k/k$. Fix $k$ and write $n=qk+r$, $0\le r<k$. ThenThus $\mathrm{limsup}{x}_n/n\le \alpha$, while the definition gives $\mathrm{liminf}{x}_n/n\ge \alpha$. Hence $x_n/n\to \alpha$.