Let the radii of the circles centered at $A, B, C$ be $x, y, z$ respectively. We simply need the existence of a point $X$ inside or on the sides of $ABC$ with the following system of inequalities holding true : \begin{align*} AX \le s-x \\ BX \le s-y \\ CX \le s-z \end{align*}Thus we construct discs $\mathcal{C}_a, \mathcal{C}_b, \mathcal{C}_c$ with centers $A, B, C$ and radii $s-x, s-y, s-z$ respectively. Now observe that we can increase the radii of the mentioned discs such that at least two pairs of them are tangent. It suffices to find a point $X$ for this case, since the older discs still remain in the circular hull. (We did this just to restrict the degrees of freedom by 2) Without loss of generality, we can assume $\mathcal{C}_a$ is the disc that is tangent to the other two. So, $y=b-x, z=c-x$. Note that the assumption leads us to the fact that $(b-x)+(c-x) \le a$, so $x \ge s-a$. This also means $x+y+z = b+c-x \le s$. We can also assume without loss of generality that $AC \ge AB$. We now claim that if the disc with center on $AC$ and radius $c-x$, and externally tangent to $\mathcal{C}_a$ is constructed, and it meets $AC$ for the second time at $M$ and the boundary of $\mathcal{C}_c$ meets line $AC$ at $N$ (outside the segment $AC$), then the midpoint of $MN$ is the required $X$. To see this, we check all inequalities one-by-one. $AX = MX - MA = (b+c-x) - 2(c-x) - x = b-c \le b-x < s-x$ $CX = NX - NC = (b+c-x) - (b-x) = c \le s-z$ (This is true since $x \ge s-a$ and $x+z=c$, so $s-z \ge c$) $BX \le AX + AB = b \le s-y $ (Analogous reasoning works as in the previous case) Since all the inequalities are verified, we are done.