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 $\omega'$ with the centre $O'$ contains the circle $\omega$ if and only if its radius is at least $O'O + r$.

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 $\omega_a \cap \omega_b \neq \emptyset$). 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.