As I did not check the details, please note that *this answer is provided with no warranty; use it at your own risks.*

If you study the functional $F(d_1,\dots,d_n)$ over the domain defined by $\sum d_i=2m$ and $d_i\geq 1$, you should find that the only critical point is the one that gives your upper bound (which, by the way, is a sibling of the arithmetico-geometric inequality and a consequence of concavity of $\sqrt{\cdot}$ : one is comparing the arithmetic mean with its $\sqrt{\cdot}$-conjugate). It follows that any minimum of $F$ must be localized on the boundary, and by symmetry we can assume $d_n=1$. Then optimizing in $d_1,\dots,d_{n-1}$ is the same game, so you should find that the minimum is obtained precisely when all but one of the $d_i$ are $1$, and the remaining one is $2m-n+1$, giving
$$\sum_1^n \sqrt{d_i} \geq \sqrt{2m-n+1}+n-1$$
and this must be achieved by any star-graph.

Note that in terms of $n$ alone, as connectedness implies $m\geq n-1$ you get
$$\sum_1^n \sqrt{d_i} \geq \sqrt{n-1}+n-1$$
(again achieved by a start graph).