Suppose $ABC$ is a triangle with $|AB|=c$, $|BC|=a$, $|CA|=b$. Suppose further that $A,B,C$ are the centers of three disks with radii $r_A,r_B,r_C$, respectively.
Is there a sensible algebraic condition (inequality?) involving $a,b,c,r_A,r_B,r_C$ equivalent to the statement "these three disks have nonempty intersection"?
Particular case when $r_a= r_B= r_C = r$. Then the quesition is when the triangle $ABC$ is contained in a circle of radius $r$. We know that the smallest circle of radius $r_{\text{min}}$ containing $\triangle ABC$ is the circumscribed on if the triangle is not obtuse and the circle of diameter the largest side it the triangle is obtuse. Now the condition is $r \ge r_{\text{min}}$.
In general the condition is semialgebraic in $a$, $b$, $c$, $r_A$, $r_B$, $r_C$, obtained by elimination, although that could be rather involved.
