jmc

The given conditions are symmetric in $a,$ $b,$ and $c,$ so without loss of generality, we can assume that $a\le b\le c.$ Then $10=a+b+c\le 3c,$ so $c\ge \frac{10}{3}.$ By AM-GM, $$(a+b{)}^2\ge 4ab.$$Then $$(10-c{)}^2\ge 4(25-ac-bc)=100-4(a+b)c=100-4(10-c)c.$$This reduces to $3c^2-20c=c(3c-20)\ge 0,$ so $c\le \frac{20}{3}.$

Now, $$m=\min \{ab,ac,bc\}=ab=25-c(a+b)=25-c(10-c)=(c-5{)}^2.$$Since $\frac{10}{3}\le c\le \frac{20}{3},$ $m=ab\le \frac{25}{9}.$

Equality occurs when $a=b=\frac{5}{3}$ and $c=\frac{20}{3},$ so the maximum value of $m$ is $\boxed{\frac{25}{9}}.$