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By GM-HM applied to 1 and $\frac{a}{b+c+d},$ $$\sqrt{1\cdot \frac{a}{b+c+d}}\ge \frac{2}{\frac{1}{1}+\frac{b+c+d}{a}}=\frac{2a}{a+b+c+d}.$$Similarly, $$\begin{array}{rl}\sqrt{\frac{b}{a+c+d}}& \ge \frac{2b}{a+b+c+d},\\ \sqrt{\frac{c}{a+b+d}}& \ge \frac{2c}{a+b+c+d},\\ \sqrt{\frac{d}{a+b+c}}& \ge \frac{2d}{a+b+c+d}.\end{array}$$Adding up all these inequalities, we get $$\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{a+c+d}}+\sqrt{\frac{c}{a+b+d}}+\sqrt{\frac{d}{a+b+c}}\ge \frac{2a+2b+2c+2d}{a+b+c+d}=2.$$The only we can get equality is if $$\begin{array}{rl}a& =b+c+d,\\ b& =a+c+d,\\ c& =a+b+d,\\ d& =a+b+c.\end{array}$$Adding these equations, we get $a+b+c+d=3(a+b+c+d),$ so $a+b+c+d=0,$ which is impossible. Thus, equality is not possible.

However, by setting $a=c=1$ and $b=d=ϵ,$ where $ϵ$ is a small positive number, then $$\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{a+c+d}}+\sqrt{\frac{c}{a+b+d}}+\sqrt{\frac{d}{a+b+c}}=2\sqrt{\frac{1}{1+2ϵ}}+2\sqrt{\frac{ϵ}{2+ϵ}}.$$As $ϵ$ approaches 0, the expression approaches 2. Thus, we can make the expression arbitrarily close to 2, so $m=\boxed{2}.$