Consider the plane region $S_n$ bounded from above and below for the graphs of $f_n(x)=x^{1/n}$ and $g_n(x)=x^n$, $0\le x\le1$.
How to find the radious and center of the circle inscribed in $S_n$?
Intuitively, the center is on the set $\{(A,A):0\le A\le 1\}$, so we can think the problem finding the value of $A$ and $r$ such that the equation $(x-A)^2+(x^n-A)^2=r^2$ has only one solution for $x$.
Another approach is parametrization, but equations are not easy to solve.
Is there any hint?
I believe that there are many circles inscribed in that region. As everybody else noticed, the diagonal $y=x$ is axis of symmetry of the region. Fix $x_0\in(0,1)$. Imagine that a (circular) planar wave is born at the point $(x_0,x_0)$ on the diagonal. It evolves as a growing circle centered at $(x_0,x_0)$. At the first moment the wave touches the two curves it becomes a circle inscribed in that region.Due to symmetry it touches both graphs at the same time)
The MAPLE animation below illustrates the process. In the animation I chose $n=4$, $x_0=0,75$.
So, if you choose the center of the circle to be $(x_0,x_0)$ you are looking to find the distance from this point to the curve $y=x^n$. In other words, you want to minimize the function $$ f(x)=(x-x_0)^2+(x^n-x_0)^2. $$ The radius of the inscribed circle centered at $(x_0,x_0)$ will then be
$$ r(x_0)=\Big(\;\min_{x\in [0,1]}(x-x_0)^2+(x^n-x_0)^2\;\Big)^{\frac{1}{2}}. $$ The equation $$ f'(x)=0 \;\Longleftrightarrow 2(x-x_0)+2nx^{n-1}(x^n-x_0)=0 \tag{1} $$ describes a curve in the $(x,x_0)$ plane. The case $n=4$ is depicted below. (The parameter $x_0$ is measured on the vertical axis.)
More concretely we can solve (1) for $x_0$
$$ x_0(1+nx^{n-1})= x+nx^{2n-1} \implies x_0=\underbrace{\frac{x+nx^{2n-1}}{1+nx^{n-1}}}_{c(x)}. $$
This shows that we can predict the location of the center of the circle given the location of its tangencies with the curve $y=x^n$. For $n=10$ the graph of the function $c(x)$ is depicted below.
It shows that for $x_0$ (roughly) between $0.4$ and $0.55$ the function
$$ x\mapsto (x-x_0)^2+(x^n-x_0)^2 $$
has three critical points. The figure below illustrates what happens when $n=10$ and $x_0=0.45$. We see that it has two local minima and a local max. We depicted below the graph of this function for $n=10$ and $c=0.4$.
Remark 1. The radius of the inscribed circle has the form $$ r(x)=\sqrt{\big(x-c(x)\big)^2+\big( x^n-c(x)\big)^2}. $$ For the circle to be actually inscribed it is necessary that $r(x)\leq c(x)$.
For $n=10$ the function $r(x)$ is depicted below, in red, and the function $c(x)$, in blue.
Remark 2. The maximal radius $r_{\max}$ is the solution of the min-max problem $$ r_{\max}=\max_{y\in (0,1)}\min_{x\in (0,1)}\sqrt{F_n(x,y)}, $$
$$ F_n(x,y)=(x-y)^2+(x^n-y)^2 $$
This suggests looking at the critical points of $F_n$. They are determined by the equations
$$ (x-y)+nx^{n-1}(x^n-y)=0, \tag{2}$$
$$ y-x+y-x^n=0. \tag{3} $$
From (3) we deduce $$ y=\frac{x+x^n}{2}. $$ Using this in (2) we deduce
$$ \frac{x-^n}{2}-nx^{n-1}\frac{x-x^n}{2} =0,$$
so
$$x=\left(\frac{1}{n}\right)^{\frac{1}{n-1}}=:t_n. $$
The center is located at $(c(t_n),c(t_n))$. Recalling that $nt_n^{n-1}=1$ we deduce
$$c(t_n)=\frac{t_n\big(1+nt_n^{2(n-1)}\big)}{1+nt_n^{n-1}}=\frac{t_n(1+t_n^{n-1})}{2}=\frac{n+1}{2n}t_n.$$
For $n=10$ we deduce
$$t_n\approx 0.7742,\;\;c(t_n)\approx 0.4258 $$
and
$$ r(t_n)\approx 0.4927. $$ The circle of radius $r(t_n)$ centered at $(c(t_n),c(t_n))$ is not even contained in the first quadrant, yet it is tangent to $y=x^n$. Thus the min-max approach does not yield $r_{\max}$ as indicated in one of @ksoriano's comments; see the figure below . depicting the circle of radius $r(t_n)$ centered at $(c(t_n), c(t_n))$, $n=10$.
The next figure depicts in the case $n=10$ an inscribed circle of radius $\approx 0.45$. It is tangent to $y=x^n$ at $x=0.9$ and $c(0.9)\approx 0.46$. Note that the radius of an inscribed circle is at most $0.5$ so the situation below is nearly optimal.
It seems to me that, for large $n$, there is a unique largest inscribed circle, which has four points of tangencies with the graphs and as $n\to \infty$ its radius converges to $0.5$, the radius of the circle inscribed in the square $[0,1]\times [0,1]$.
Remark 3. . Recall that the map $x\mapsto c(x)$ determines the location of the center tangent to $y=x^n$ at the point $(x,x^n)$. Its radius is $r(x)$. The plot of the $(c, r)$ curve $\newcommand{\bR}{\mathbb{R}}$ $$ [0,1]\ni x\mapsto \big(\; c(x), r(x)\;\big)\in\bR^2, $$ is very revealing. Below we depict the case $n=10$.
The location of the center is tracked on the horizontal axis. As the center moves along the diagonal from $(0,0)$ to $(1,1)$ we notice various "phase transitions" or bifurcations. Imagine a vertical line moving from left to right in the above figure and keep track of the number of points of intersection with the $(c,r)$ curve.
Depending on the location of the center, there are typically, either one, or three circles tangent to $y=x^n$. To detect $r_{\max}$ we need to keep track of the smallest of these circles. In the above plot this means looking only at the bottom triangle of the $(c,r)$-curve. The altitude of the top vertex of this triangle is $r_\max$.
In any case, the above figure suggests that the function $$ y\mapsto \min_{x\in[0,1]} F_n(x,y) $$ is rather regular.
There exist numbers $y_{\min}<y_{\max}$ in $(0,1)$ so that for $y\in[0,1]\setminus [y_{\min},y_{\max}]$ there exists only one circle centered at $(y,y)$ and tangent to the boundary of the region. $\newcommand{\pa}{\partial}$
As $y$ crosses $y_{\min}$ increasingly the equation $$ \pa_x F_n(x,y)=0 $$ undergoes a qualitative change: it goes from having one solution to having three solutions. It is a "birth" process.
As $y$ crosses $y_{\max}$ increasingly, the above equation goes from having three solutions to having one solution. It is a "death" process.
The points $y_{\min}$ and $y_{\max}$ are determined by solving the nonlinear system $$ \pa_x F_n(x,y)=\pa^2_{xx}F_n(x,y)=0, $$ i.e., $$ nx^{2n-1}-nx^{n-1}y+x-y=0,\;\;n(2n-1)x^{2n-2}+n(n-1)x^{n-2}y+1=0. $$ I believe that for $n$ large this system has three solutions $$ (x_{\min},y_{\min}),\;\;(x_*,y_*),\;\;(x_{\max},y_{\max}), $$ $$ y_{\min}<y_*<y_{\max}. $$ The center of the larges inscribed circle is located $(y_*,y_*)$ and its radius is $$ r_{\max}=r(x_*). $$ The number of solutions of the system is the number of roots in $(0,1)$ of the equation $$ (2n-1)x^{2n-2}+(n-1)x^{n-2}c(x)+\frac{1}{n}=0. $$ This reduces to a polynomial equation. Theoretically, the number of solutions of a real polynomial equation in a given interval can be determined using the classical Sturm theorem.
If the curves constitute to a function and its inverse, as is the case in your example, then you have to check pairs of points $p:=(x_a,f(x_a)),\ q:=(x_b,f^{-1}(x_b))=(f(x_a),x_a)$, so that:
$$\frac{d}{dx}f(x_a)=1=\frac{d}{dx}f^{-1}(x_b)\quad \wedge\quad \frac{f^{-1}(x_b)-f(x_a)}{x_b-x_a}=-1$$
if, as is the case in your example, the region defined by the two curves is also convex, that pair of points is unique and thus $(p,q)$ marks a diameter of the largest contained circle and thus $p=((\frac{1}{n})^\frac{1}{n-1},(\frac{1}{n})^{\frac{n}{n-1}}),\ q=((\frac{1}{n})^{\frac{n}{n-1}},(\frac{1}{n})^\frac{1}{n-1})$ define the pair of points in which the largest contained circle touches both curves; the diameter equals the distance of $p$ and $q$
