Difference of maximum functions over finite alternatives

Question

I wish to find an upper bound on $c$, where $c$ is defined as

$$c = \max\{a_1,a_2,a_3\} - \max\{b_1,b_2,b_3\}$$

Is it true that $c\le\max\{a_1-b_1,a_2-b_2,a_3-b_3\}$?

Answer 1

Yes, it's true that with

$$c = \max\{a_1,a_2,a_3\} - \max\{b_1,b_2,b_3\} \tag{1}\label{eq1A}$$

you always get

$$c \le \max\{a_1-b_1,a_2-b_2,a_3-b_3\} \tag{2}\label{eq2A}$$

Suppose that $\max\{a_1,a_2,a_3\} = a_1$. Then if $a_1$ is subtracted by anything less than $\max\{b_1,b_2,b_3\}$, it would be larger than $c$. Thus, $a_1 - b_1 \ge c$, so this means \eqref{eq2A} holds. You can use a similar argument for the cases where $a_2$ or $a_3$ are the maximum instead to show \eqref{eq2A} holds in those cases as well.