Let $S = \{x\in [1,\infty) : f(x) < \sqrt{2x}\}$. Note that from $f(2x) \leq f(x)\sqrt{2}$, we get that if $x\in S$, then $2x\in S$. So it is enough to prove the statement for $x\in[1,2)$. But that is obvious since if $1\leq x < 2$, we have $f(x) \leq x < \sqrt{2x}$ so we are done.