Min max max

When $x \ge 3\implies 5-\max(3,x) = 5-x < 3+x \implies \max(5-\max(3,x), 3+x) = 3+x \ge 6$ .

When $x \le 3\implies \max(3,x) = 3$ and $5-\max(3,x) = 5-3=2$ .

When $x \le -1\implies 3+x \le 2$ and $\max(5-\max(3,x), 3+x) = \max(2, \le 2)=2$ .

$\implies \max(5-\max(3,x), 3+x) = \begin{cases} 2 & \text{for } x \le -1 \\ 3+x > 2 & \text{for } x > -1 \end{cases}$

Therefore, the minimum value of $\max(5-\max(3,x), 3+x)$ is $\boxed{2}$ .