# Prove that the $g((a_1,b_1),(a_2,b_2))=max\{|a_1-a_2|,|b_1-b_2|\}$ is a metric?

I have been asked to prove the following but I am unsure how to show that the triangle inequality is true for this function.

If $$g((a_1,b_1),(a_2,b_2))=max\{|a_1-a_2|,|b_1-b_2|\}$$, show that g is a metric?

Since \begin{aligned} |a_1-a_3| &= |a_1-a_2+ a_2 -a_3|\\ &\le |a_1-a_2| + |a_2-a_3|\\ &\le g((a_1,b_1), (a_2,b_2)) + g((a_2,b_2), (a_3,b_3)). \end{aligned} Similarly $$|b_1-b_3| \le g((a_1,b_1), (a_2,b_2)) + g((a_2,b_2), (a_3,b_3)).$$ So $$g((a_1,b_1),(a_3,b_3)) \le g((a_1,b_1), (a_2,b_2)) + g((a_2,b_2), (a_3,b_3)).$$