HMMT February

Without loss of generality, we assume that $c \geq b \geq a$. We see that $3 c \geq a+b+c=10$. Therefore, $c \geq \frac{10}{3}$. Since $0 \leq(a-b)^{2} =(a+b)^{2}-4 a b =(10-c)^{2}-4(25-c(a+b)) =(10-c)^{2}-4(25-c(10-c)) =c(20-3 c)$ we obtain $c \leq \frac{20}{3}$. Consider $m=\min \{a b, b c, c a\}=a b$, as $b c \geq c a \geq a b$. We compute $a b=25-c(a+b)=25-c(10-c)=(c-5)^{2}$. Since $\frac{10}{3} \leq c \leq \frac{20}{3}$, we get that $a b \leq \frac{25}{9}$. Therefore, $m \leq \frac{25}{9}$ in all cases and the equality can be obtained when $(a, b, c)=\left(\frac{5}{3}, \frac{5}{3}, \frac{20}{3}\right)$.