min $\left[{\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}^2+{\left(3+\sqrt{1-{\mathrm{x}}_1^2}-\sqrt{4{\mathrm{x}}_2}\right)}^2\right],\mathrm{\forall }{\mathrm{x}}_1,{\mathrm{x}}_2\in \mathrm{R}$,is 

Let ${\mathrm{y}}_1=3+\sqrt{1-{\mathrm{x}}_1^2}$ and ${\mathrm{y}}_2=\sqrt{4{\mathrm{x}}_2}$
or ${\mathrm{x}}_1^2+{\left({\mathrm{y}}_1-3\right)}^2=1$ and ${\mathrm{y}}_2^2=4{\mathrm{x}}_2$
Thus $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ is  on the circle ${\mathrm{x}}^2+(\mathrm{y}-3{)}^2=1$.
Also, $\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ lies on the  upper half of the  parabola ${\mathrm{y}}^2=4\mathrm{x}$.
Thus,  the  given  expression  is square  of the shortest  distance between the curves ${\mathrm{x}}^2+(\mathrm{y}-3{)}^2=1$ and ${\mathrm{y}}^2=4\mathrm{x}$.
Now,  the shortest distance  always  occurs  along  the  common normal  to the curves  and  the normal to the circle passes  through the center of the  circle.
Normal  to the  parabola  ${\mathrm{y}}^2=4\mathrm{x}$ is $\mathrm{y}=\mathrm{mx}-2\mathrm{m}-{\mathrm{m}}^3$. lt passes through  (0, 3). Therefore,${\mathrm{m}}^3+2\mathrm{m}+3=0$, which  has  only  one  root,$\mathrm{m}=-1$. Hence, the corresponding  point on  the  parabola  is (1,2). Thus, the  required  minimum  distance is $\sqrt{1+1}-1=\sqrt{2}-1$.