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!  
 A: $\require{cancel}$
I'd like to elaborate on (and slightly critique) the answer given by Marty Cohen to illustrate one thing to the reader. It is incorrect to deduce from what you want to prove to prove said statement. In Marty's answer, he deducing from the line $$d_{t}(x,y)\geq d_{e}(x,y)$$
which will give the result:
\begin{align}
d_{t}(x,y)&\geq d_{e}(x,y)\\
|x_{1}\text{−}y_{1}|+|x_{2}\text{−}y_{2}|&\geq \sqrt{(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}}\\
\left(|x_{1}\text{−}y_{1}|+|x_{2}\text{−}y_{2}|\right)^{2}&\geq (x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}\\
(|x_{1}\text{−}y_{1}|+|x_{2}\text{−}y_{2}|)\cdot(|x_{1}\text{−}y_{1}|+|x_{2}\text{−}y_{2}|)&= \\|x_{1}-y_{1}|^{2}+2|x_{2}-y_{2}|\cdot|x_{1}-y_{1}|+|x_{2}-y_{2}|^{2}&= \\
\cancel{(x_{1}-y_{1})^{2}}+2|x_{2}-y_{2}||x_{1}-y_{1}|+\cancel{(x_{2}-y_{2})^{2}}&= \cancel{(x_{1}-y_{1})^{2}}+\cancel{(x_{2}-y_{2})^{2}}\\
2|x_{2}-y_{2}||x_{1}-y_{1}|&\geq 0
\end{align}
This implies the answer you want, but the logic is faulty. The correct manner to prove such statements is as follows:
\begin{align}
 d_{e}(x,y) & = \sqrt{(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}}\\
d_{e}^{2}(x,y) & = (x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2} \\
 &= |x_{1}-y_{1}|^{2}+|x_{2}-y_{2}|^{2}\\
 & \leq |x_{1}-y_{1}|^{2}+2|x_{2}-y_{2}|\cdot|x_{1}-y_{1}|+|x_{2}-y_{2}|^{2}\\
 & = (|x_{1}\text{−}y_{1}|+|x_{2}\text{−}y_{2}|)\cdot(|x_{1}\text{−}y_{1}|+|x_{2}\text{−}y_{2}|)\\
 & = \left(|x_{1}\text{−}y_{1}|+|x_{2}\text{−}y_{2}|\right)^{2}\\
 & = d_{t}^{2}(x,y) \\
\therefore d_{e}^{2}(x,y) & \leq d_{t}^{2}(x,y) \\
\implies d_{e}(x,y) &\leq d_{t}(x,y)\\
\end{align}
Notice how I don't assume anything about the relationship between $d_{t}(x,y)$ and  $d_{e}(x,y)$.
A: 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.
A: The case for $n=2$ is
$\sqrt{a^2+b^2}
\le |a|+|b|
$.
Squaring both sides,
this is
$a^2+b^2
\le a^2+2|a|\ |b|+b^2
$,
which is obviously true.
The case for general $n$ is
$\sqrt{\sum_{i=1}^n a_i^2}
\le \sum_{i=1}^n |a_i|
$.
If we square both sides of this,
we get
$\sum_{i=1}^n a_i^2
\le (\sum_{i=1}^n |a_i|)^2
$.
But
$(\sum_{i=1}^n |a_i|)^2
=\sum_{i=1}^n |a_i|^2
+2\sum_{i=1}^n \sum_{j=1}^{i-1} |a_i|\ |a_j|
\ge\sum_{i=1}^n a_i^2
$.
A: Here's another way.
Let $$z_0 = x = (x_1,\dots,x_n) \\z_1 = (y_1,x_2,x_3,\dots,x_n) \\ \dots\\ z_j =(y_1,\dots,y_j,x_{j+1},x_{j+2},\dots,x_n)\\ \dots \\ z_n = (y_1,\dots,y_n).$$ 
Then the inequality you want to prove is exactly
$$d^2(z_0,z_n) \leq d^2(z_0,z_1) + d^2(z_1,z_2) + \dots + d^2(z_{n-1},z_n).$$
But this follows from the triangle inequality for $d^2$.
