Show that the Euclidean Metric is less than or equal to the Taxi-cab metric for $\mathbb{R}^{n}$ I am trying to prove that the Euclidean metric; $(\mathbb{R}^{n},d^{2})$; defined: $$d^2(x,y) = \sqrt{\sum_{i=1}^n(x_i-y _i)^2}. $$ is less than or equal to the Taxi Cab metric; $(\mathbb{R}^{n},d^{1})$ defined: $$d^1(x,y) = \sum_{i=1}^{n}|x_i-y_i|.$$  It is obvious to me that this like comparing the length of a hypotenuse with the length of two sides of a triangle. The only thing I can think of to illustrate this inequality is that if I observe that in each case, $d: X \times X \rightarrow \mathbb{R}$ I have that for $d^2(x,y)= \sqrt{r^2}$ and $d^{1}(x,y) = |r|$, clearly $$\sqrt{r^2} \le |r|.$$ This is suspiciously too simple. For example, it seems that it is missing the scenario when the L.H.S. is less than the R.H.S., but is that necessary? Any tips are appreciated. Thanks!  

If you try squaring both sides of your inequality, you'll get something which obviously holds: all terms on the left will appear on the right, plus more, and all terms will be positive.
