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To simplify the formulations, we say that a point lies inside of the circle if it lies on that circle or in its interior. Assume we are given a circle $\omega$ with the radius of $r$ and the centre $O$. A circle ${\omega }^{\prime }$ with the centre $O^{\prime }$ contains the circle $\omega$ if and only if its radius is at least $O^{\prime }O+r$.



Denote by $r_a$, $r_b$, $r_c$ the radii of our circles with the centres at $A$, $B$ and $C$, respectively. Using our observation three times indicates that the centre $X$ of the circle we are seeking has to meet $s\ge AX+r_a$, or equivalently $AX\le s-r_a$, and analogously $BX\le s-r_b$ and $CX\le s-r_c$.

Notice that the numbers $s-r_a$, $s-r_b$ and $s-r_c$ are positive. We will show this for $s-r_a$. Since our circles are disjoint with disjoint interiors, we know that $r_a<b$ and $r_a<c$. This gives us $r_a<(b+c)/2<(a+b+c)/2=s$, which indeed means that $s-r_a$ is a positive number.

Now we may consider three circles with the centres $A$, $B$ and $C$ and radii $s-r_a$, $s-r_b$ and $s-r_c$, respectively. If we prove that there is a point $X$ lying inside each of them, we will be done.

Each two of these three circles intersect at two points, because for example $(s-r_a)+(s-r_b)>2s-c=a+b>c$ (and also $c>|(s-r_a)-(s-r_b)|$). For the sake of contradiction assume there is no point lying inside all of them. Then the situation looks like on the picture, that is, there exists a point $X$ inside of the triangle which lies outside of the three circles (see the remark at the end):



For such $X$ we have $AX+BX+CX>s-r_a+s-r_b+s-r_c>2s$. This is not possible, however. Let $Y$ be the intersection of $BX$ and $AC$. Then using the triangle inequalities for the triangles $CXY$, $ABY$ we get $$BX+CX<BX+XY+CY=BY+CY<AB+AY+CY=AB+AC.$$

Similarly $AX+BX<AC+BC$ and $CX+AX<BC+AB$. Summing these three inequalities we obtain $AX+BX+CX<AB+BC+AC=2s$, which is a contradiction.

Remark. If three circles ${\omega }_a$, ${\omega }_b$ and ${\omega }_c$ with the centres $A$, $B$ and $C$, respectively, satisfy the conditions that each two of them intersect and there is no point lying inside all of the three circles, then there exists a point in the interior of the triangle $ABC$ which lies outside of each of the three circles.



To prove this, consider the intersection point $P$ of ${\omega }_b$ and ${\omega }_c$ which lies in the halfplane determined by the line $BC$ and the point $A$. The intersection $Q$ of the ray $BA$ with ${\omega }_b$ lies inside of ${\omega }_a$, since it is the closest point of ${\omega }_b$ to $A$ (this is true even if $A$ is inside of ${\omega }_b$, since ${\omega }_a\cap {\omega }_b\ne \varnothing$). Therefore $A$ cannot lie in the angle $CBP$ (otherwise $Q$ would lie inside all of the three circles). But that means $P$ lies in the interior of the angle $CBA$. Similarly $P$ lies in the interior of the angle $BCA$. So we have that $P$ lies in the interior of the triangle $ABC$. Since $P$ does not lie inside of ${\omega }_a$, there is a point in the neighbourhood of $P$ lying outside of all the three circles.