Assuming that we have an algorithm that decides SAT in $2^{O(\sqrt{n})}$, can every language in NP be decided in that time?
I had the following idea:
Because SAT is NP-complete, every language in NP can be reduced to it using a polynomial-time reduction. Therefore, every language in NP can be decided in $2^{O(\sqrt{n})} + O(polynomial)$ (the time that is required for the reduction). Because $2^{O(\sqrt{n})}$ grows asymptotically faster than $O(polynomial)$, every language in NP is decidable in $2^{O(\sqrt{n})}$.
Is that a proper explanation? I'm not quite sure about the last sentence.