smc

Since $s=AC+BC$, $s^2=A{C}^2+2\cdot AC\cdot BC+B{C}^2$. Since $△ABC$ is inscribed and $AB$ is the diameter, $△ABC$ is a right triangle, and by the Pythagorean Theorem, $A{C}^2+B{C}^2=A{C}^2=(2r{)}^2$. Thus, $s^2=4r^2+2\cdot AC\cdot BC$. The area of $△ABC$ is $\frac{AC\cdot BC}{2}$, so $2\cdot [ABC]=AC\cdot BC$. That means $s^2=4r^2+4\cdot [ABC]$. The area of $△ABC$ can also be calculated by using base $AB$ and the altitude from $C$. The maximum possible value of the altitude is $r$, so the maximum area of $△ABC$ is $r^2$. Therefore, $s^2\le 8r^2$, so the answer is $\boxed{s^2\le 8r^2}$.