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Consider an equilateral triangle $ABC$ with side lengths 1, on the picture with its circumcircle outlined. Its circumradius will be $1/sqrt{3}$.

Now imagine we allow each vertex to move within a disc of radius $rho$ centered at that vertex. We end up with a new triangle $A'B'C'$, where e.g. $A' in B(A,rho)$, the disc with center $A$ and radius $rho$.

The question is simple:

What is the maximum circumradius of the perturbed triangle $A'B'C'$?

According to Existence of Gibbsian point processes with geometry-dependent interactions (though I wouldn't recommend looking through the paper for any insights, as it deals with a completely different topic and gives no details on this problem) the maximum circumradius, for $rho leq sqrt 3 /6$, is
$$1/sqrt{3} + rho$$
Which is a very simple solution, but though it intuitively makes some sense to me, I can't really convince myself of it or prove it. It also leads to a secondary question

What happens at $rho = sqrt3/6$? How does the solution change for $rho > sqrt3 / 6$?

Intuitively what I think happens is that for small enough $rho$ the solution is still an equilateral triangle but after a certain point (probably $rho = sqrt 3/6$) this is no longer the case, as moving the points closer to collinearity will yield a greater circumradius, until finally at $rho = sqrt3/4$ the points can become collinear.

Why the circumradius is maximised when all three perturbed points are on the boundary of the small circles

Suppose the small circles are disjoint, there is no line intersecting all three circles, points $A'$ and $B'$ are fixed and $C'$ is not on the boundary of its circle (i.e. it has a neighbourhood within the circle around $C$). The circumradius is a differentiable, non-constant function of the coordinates of $C'$. Thus there must be some perturbation of $C'$, within the aforementioned neighbourhood, that increases the circumradius – there are no local extrema to impede this modification because the level sets of the circumradius function can only be as small as the circle with diameter $A'B'$, whereas arbitrarily small level sets can be drawn around a local extremum.

Thus the circumradius can only be maximised when $A',B',C'$ are all on the boundaries of their circles, at which point the circumcircle is tangent to all three small circles.

Now the problem boils down to determining which of two such circles – $Gamma_3$ internally tangent to all three small circles, or $Gamma_2$ internally tangent to only two – is larger:

(It can be easily shown that the corresponding $Gamma_1$ and $Gamma_0$ are never larger than $Gamma_3$; we choose the vertices of $triangle A'B'C'$ as the points of tangency.) It is easy to show that $r(Gamma_3)=frac1{sqrt3}+rho$, since it is concentric with the original equilateral triangle, which has circumradius $frac1{sqrt3}$. To determine the radius of $Gamma_2$, we draw another diagram:

From this we derive the equation
$$frac{sqrt3}2-rho+k-rho=sqrt{frac14+k^2}$$
$$(k+sqrt3/2-2rho)^2=k^2+(sqrt3-4rho)k+(sqrt3/2-2rho)^2=frac14+k^2$$
$$(sqrt3-4rho)k+frac34-2sqrt3rho+4rho^2=frac14$$
$$k=frac{-frac12+2sqrt3rho-4rho^2}{sqrt3-4rho}$$
$$r(Gamma_2)=frac{-frac12+2sqrt3rho-4rho^2}{sqrt3-4rho}+frac{sqrt3}2-rho=frac{1-sqrt3rho}{sqrt3-4rho}$$
Now where is this greater than $r(Gamma_3)$?
$$frac{1-sqrt3rho}{sqrt3-4rho}=frac1{sqrt3}+rho$$
$$1-sqrt3rho=(1/sqrt3+rho)(sqrt3-4rho)=1-4/sqrt3rho+sqrt3rho-4rho^2$$
$$4sqrt3rho^2-2rho=0qquadrho=frac{sqrt3}6$$
Thus, above this critical value $Gamma_2$ is larger and vice versa.

The vertical asymptote of $r(Gamma_2)=frac{1-sqrt3rho}{sqrt3-4rho}$ comes at $rho=frac{sqrt3}4$; at and above this value we can choose three collinear points.

Thus the maximum circumradius of $triangle A'B'C'$ is
$$begin{cases}frac1{sqrt3}+rho&rho<frac{sqrt3}6\
frac{1-sqrt3rho}{sqrt3-4rho}&frac{sqrt3}6lerho<frac{sqrt3}4\
infty&rhogefrac{sqrt3}4end{cases}$$