 Solution

Fix $k\ge 1$ and write $n=qk+r$ with $1\le r\le k$ and $q\ge 0$. Repeatedly use the given inequality with its second index equal to $k$. This yieldsThere are only finitely many possible remainders for fixed $k$, soTaking $k=1$ rules out positive infinity for this upper limit. Let $L=\mathrm{liminf}{x}_n/n$, which may be negative infinity, and choose $k_j\to \infty$ with $x_{k_j}/k_j\to L$. The assumption ${\alpha }_{k_j}/k_j\to 0$ then gives $\mathrm{limsup}{x}_n/n\le L$. This also works when $L=-\infty$, by taking an arbitrarily negative upper bound. Therefore the asymmetrically almost-subadditive sequence hasThe last infimum identity follows from the fixed-$k$ bound and from the convergence of the corrected ratios to $\gamma$. No positivity assumption on ${\alpha }_n$ is required.