Q: how to get rid of the denominator in the squareroot, that appears in the denominator? Here is an example:  $\frac{\nu  y_0}{\left(x_0+\frac{1}{2}\right) \sqrt{\nu ^2+\frac{\left(4 x_0^2+4 y_0^2-1\right){}^2}{16 y_0^2}}}$
 test=(\[Nu] Subscript[y, 0])/((1/2+Subscript[x, 0]) Sqrt[\[Nu]^2+(-1+4 Subsuperscript[x, 0, 2]+4 Subsuperscript[y, 0, 2])^2/(16 Subsuperscript[y, 0, 2])])

I'd like to get rid of the denominator in the squareroot. I tried separating the numerator and denominator, and multiply both by $4y_0$
 or $\sqrt{16y_0^2}$. Somehow neither works. Strangely, even if I multiply the denominator by $\sqrt{16y_0^2}$, the two terms won't combine together even if I use Simplify.
Currently I use FullSimplify[test,Subscript[y, 0]>0], which does not work. Hence, I had to copy the material in the squareroot, multiply by $16y_0^2$, simplify it  separately, then plug back the original expression to obtain the results (the numerator will be multiplied by $4y_0$, too). 
I wonder if there is any better way to do this mor efficiently? 

A: Try this:
test = (ν y)/((1/2 + 
      x) Sqrt[ν^2 + (-1 + 4 x^2 + 4 y^2)^2/(16 y^2)]);

test2 = MapAt[Together[#] &, test, {4, 1}]
test3 = MapAt[Hold, test2, {5, 1, 2}] // Simplify[#, y > 0] & //ReleaseHold


Have fun!

A: Or:
test = (\[Nu] y)/((1/2 + 
   x) Sqrt[\[Nu]^2 + (-1 + 4 x^2 + 4 y^2)^2/(16 y^2)]);

test // Together // 
        PowerExpand[#, Assumptions -> y \[Element] Reals && y != 0] &


