Let $N\geq 1$ be a positive integer and assume that $N=2^{n_1}+2^{n_2}+\cdots+2^{n_{p}}$, $n_{1}>n_{2}>\cdots>n_{p}\geq 0$, is the binary representation of $N$. I believe that the following inequality is valid for every $0<s<1$ but I haven't been able to prove it: $$ \left(\frac{2^{n_1}}{N}\right)^s+\left(\frac{2^{n_2}}{N}\right)^s+\cdots+\left(\frac{2^{n_{p}}}{N}\right)^s\leq \frac{1}{2^s-1}. $$ Note that by continuity the inequality is valid for $s$ sufficiently close to zero since the left side approaches $p$ and the right side approaches infinity. When $s=1$ we get equality.

Note that one can assume that $N$ is odd, i.e., $n_p=0$. If $N$ is of the form $N=1+2+2^2+\cdots+2^{p-1}=2^{p}-1$, then one obtains a geometric sum and the inequality reduces to $2^{sp}-1\leq (2^p-1)^s$, which follows from $(a+b)^s\leq a^s+b^s$.