Question: $y={\dfrac{[3x+1]}{[4x-1]^\frac{1}{2}}}$

Is it logical to square both the numerator and denominator to cancel out the power to $\frac12$ in the denominator?

Edit: This expression is being used for a chain rule application.

Answer: 
This can be done, but unless someone has a bright idea, the algebra is pretty revolting. Start with
$$\eqalign{
 x^2+y^2=r^2&\qquad\qquad(1)\cr
 (x-1)^2+y^2=(r+b)^2&\qquad\qquad(2)\cr
 (x-\tfrac12)^2+(y-\tfrac{\sqrt3}2)^2=(r+a)^2&\qquad\qquad(3)\cr}$$
If you now take $(a-b)$ times $(1)$, minus $a$ times $(2)$, plus $b$ times $(3)$, then assuming my algebra is correct you get
$$(2a-b)x-b\sqrt3\,y=(a-b)(ab+1)\ .$$
Now solve for $y$ in terms of $x$; use $(1)$ to get $r$ in terms of $x$; then use these and $(2)$ to get an equation in $x,a,b$ only. If, again, my algebra is correct (highly unlikely), you end up with the quadratic
$$(12-12b^2-16a^2+16ab-4b^2)x^2+(-12+12b^2+8(2a-b)(a-b)(ab+1))x+(3+3b^4-6b^2-4(a-b)^2(ab+1)^2)=0\ ,$$
which hopefully will turn out to have real solutions.

Sorry for the cop-out but I am definitely leaving the rest to you :)
Let the coordinates of the triangle be $(0,0),(1,0),\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$. Then, we can write three equations of the form

$$(x-x\_0)^2+(y-y\_0)^2=r\_0^2$$

for various $x\_0,y\_0,r\_0$. What happens if you assume a solution to each of these concurrently and then solve for $r$?