Radius of a largest circle inscribed under $y=\frac{1}{(1+x^2)^n}$, closed form The curve $y=\frac{1}{1+x^2}$ has an obvious connection to circles, because it's the derivative of the arctangent function.
Besides, if we inscribe a circle under it, its radius is exactly $R=\frac{1}{2}$, so it takes exactly one fourth of the full area under the curve.

Actually, I don't know an easy way to find $R$ even in this simple case. So, let's consider how I solve the more general problem.

Find the largest circle fitting between the curve $y=y (x)$ and the line $y=0$
$$y(x)=\frac{1}{(1+x^2)^n}~~~~~~ n=1,2,3,\dots$$

The distance from the circle origin $(0,R)$ to the curve is the minimum of a function:
$$s(x)=\sqrt{x^2+\left(R-\frac{1}{(1+x^2)^n} \right)^2}$$
$$s(x)'=0$$
The condition for the inscribed circle is $s(x)=R$.
Let's denote:
$$t=1+x^2$$
Then from the above we obtain the system of equations:

$$t^{2n+1}+2n~R~t^n-2n=0$$
$$t^{2n+1}-t^{2n}-2~R~t^n+1=0$$

This is how I got the solution for $n=1$ (using Mathematica). Other solutions do not have obvious closed forms. Here are the numerical roots I've got with Mathematica:
$$R_2=0.4735710971151933$$

$$R_3=0.4401444298014721$$
$$R_4=0.41216506385826285$$
And so on.
The questions I ask:

Can there be closed forms for $R$ for $n \neq 1$? How to find them?
Is there an easier way to find $R$? At least for some $n$?

Edit
The most simple equation for $t$, as far as I can see, is:

$$(n+1)~t^{2n+1}-n~t^{2n}-n=0$$

I think I might ask a separate question about this equation. It's easy to solve by Newton-Raphson, but can it have closed form solutions for any $n$?
 A: It is possible to find tangent points and the circle of contact for given $n$  using the conventional and classical Lagrangian multiplier method. The circle can be the object, and given curve the constraint function or vice-versa. A sort of non-linear programming in Operations Research. 
Obtained the same contact radii values for each $n$ as the OP.
$R = \lambda $ parameter circles with x-axis tangent at origin are
$$ F = x^2 + (y-\lambda)^2 = x^2 +y^2 -2 y \lambda = 0 \tag{1} $$
$$ y = 1/(1+x^2)^n ;\, \rightarrow G = y (1+x^2)^n - 1 =0 \tag{2} $$
We have a constant Lagrange multiplier $$ \frac{F_x}{F_y}= \frac{G_x}{G_y} = C, \; \tag{3} $$
which after algebraic simplification yields the required link relation:
$$ 2 n y (y-x) = ( 1 + x^2) \; \tag{4} $$
Solving together (1), (2) and (4) we get $ x,y, \lambda = R $  that allow the curves and maximum radius contacts to be sketched as follows for $n = 1,3,6,20.$
The radius cannot be fully expressed in closed form , but only as an algebraic number in a field given from Mathematica for $ n\ne 1 $ example:
    n=3, \; \lambda = R =  AlgebraicNumber[
    Root[-8192 + 10240 #1^2 + 9984 #1^4 + 5120 #1^6 + 1520 #1^8 + 
    264 #1^10 + 25 #1^12 + #1^14 &, 2], {1/6, 0, 5/12, 0, 1/4, 0, 11/
    192, 0, 7/1536, 0, 0, 0, 0, 0}]

EDIT 1
(deleted)

A: The radius of curvature of a function $f(x)$ is
$$ \rho =- \frac{ \left( 1 + \left(\frac{{\rm d}}{{\rm d}x}f(x)\right)^2 \right)^\frac{3}{2} } { \frac{{\rm d}^2}{{\rm d}x^2} f(x) } $$
NOTE: If the curve is given implicitly then $$\rho = \frac{ (x'^2+y'^2)^\frac{3}{2}}{y' x'' - y'' x'}$$
Case 1
Consider $f(x) = \frac{1}{1+x^2}$
$$ \rho = -\frac{ \left(1+\frac{4 x^2}{(1+x^2)^4} \right)^\frac{3}{2}} { \frac{2(3 x^2-1)}{(1+x^2)^3} } $$
and for $x=0$ the circle is $\rho =\frac{1}{2}$ as you expect.
Case 2
Consider $f(x) = \frac{1}{(1+x^2)^n}$
The radius of curvature is
$$ \rho = \frac{ \left( 4 n^2 x^2 + (1+x^2)^{2(n+1)}\right)^\frac{3}{2} }{ \tfrac{2 n (1-x^2 (2n+1))}{(1+x^2)^{-(2n+1)}}} $$
and for $x=0$ the curvature is simply ${\rho = \frac{1}{2n}}$

Now the circle is going to be tangent to the curve (one of the curvature circles) and its center is going to be located at
$$ x_c = x - \frac{ y' (1+y'^2)}{y''} \\ y_c = y + \frac{1+y'^2}{y''} $$
here $y' = \frac{{\rm d}}{{\rm d}x} f(x)$ and $y'' = \frac{{\rm d}^2}{{\rm d}x^2} f(x)$ 
For the circle to be tangent to $y=0$, its center has to be at $y_c = \rho$ or
$$ y + \frac{1+y'^2}{y''} = - \frac{(1+y'^2)^\frac{3}{2}}{y''} $$
For $y=\tfrac{1}{1+x^2}$ my CAS says the solution (besides $x=0$) is the roots of
$$ 2 x^{16} + 15 x^{14} + 48 x^{12} + 97 x^{10} + 126 x^8+81 x^6-20 x^4 -33 x^2 = 12$$ 
$ x = 0.759119999241623 $ or $\rho = 4.32460278011942$, but that ends up being a circle over the curve.

For the general case of $y = \frac{1}{(1+x^2)^n}$ the equation to be solved is
$$ \left( (1+x^2)^{2(n+1)} + 4 n^2 x^2\right)^\frac{3}{2} + (1+x^2)^{3(n+1)} + 2 n ( 1+x^2)^{n+1} \left( 4 n x^2 + x^2 -1 \right) = 0 $$
