jmc

A real number $x$ is in the domain of $g$ if and only if $$(x-3{)}^2-(x-8{)}^2\ge 0.$$ Expanding this out and simplifying, we get $$10x-55\ge 0;$$ the smallest solution is $x=\frac{55}{10}=\boxed{\frac{11}{2}}$.

Alternatively, once we have the quadratic equation $$(x-3{)}^2-(x-8{)}^2\ge 0,$$ instead of expanding it out, we can observe that $(x-3{)}^2$ is the square of the distance from $x$ to $3$ on the number line, while $(x-8{)}^2$ is the square of the distance from $x$ to $8$. Thus, $(x-3{)}^2-(x-8{)}^2\ge 0$ is true if $x$ is closer to $8$ than to $3$, which is true if and only if $x\ge \frac{8+3}{2}=\boxed{\frac{11}{2}}$.