Show the Euclidean metric and maximum metric are strongly equivalent.

I need to show that the Euclidean metric and maximum metric (or square metric??) are strongly equivalent. I have no idea how to start this proof. Any help?

$d_1, d_2$ are called strongly equivalent if there exist positive constants $K, M$ such that for all $x, y\in X$: $Md_1(x,y)\leq d_2(x,y)\leq Kd_1(x,y)$

• What does strongly equivalent mean? – Umberto P. Mar 11 '15 at 14:26
• I added the definition to the question – Shannon Mar 11 '15 at 14:29
• Similar to the definition of equivalence – Shannon Mar 11 '15 at 14:29

Work in $\mathbb R^2$ since the idea carries over easily to higher dimensions. Let $(x,y) \in \mathbb R^2$ and assume (with no loss of generality) that $|x| \le |y|$.
Since $|x|^2 + |y|^2 \le 2|y|^2$ you have $$\sqrt{x^2 + y^2} \le \sqrt 2 |y| = \sqrt 2 \max \{|x|,|y|\}.$$
Since $|x| \le |y|$ you have $$\max\{|x|,|y|\} = |y| = \sqrt{y^2} \le \sqrt{x^2 + y^2}.$$
• Thanks! I actually have to prove it in $\mathbb R^m$ so I can understand it from this. – Shannon Mar 11 '15 at 14:46