APX Hardness implies no QPTAS?
APX-Hardness implies that there is a $\delta &gt; 0$ such that the problem does not admit a $(1+\delta)$-approximation unless $P=NP$. Clearly this rules out a PTAS (assuming $P \neq NP$). As for QPTAS, it will rule it out unless you believe that NP is contained in quasi-polynomial time. As far as I know, that is the only reason why APX-Hardness rules out a QPTAS. 

Since a couple of people asked more details, here are some more. APX-Hardness for a minimization problem P implies that there is a fixed $\delta &gt; 0$ and a polynomial-time reduction from 3-SAT to P such that a $(1+\delta)$-approximation for P allows one to decide whether the 3-SAT formula is satisfiable or not. If there is a QPTAS for P we can obtain for any fixed $\delta &gt; 0$ a $(1+\delta)$-approximation in time say $n^{O(\log n)}$. But this implies that we can decide whether a 3-SAT formula is satisfiable in $n^{O(\log n)}$ time which in turn implies that NP is in QP.