§ L1 norm is greater than or equal to L2 norm

Pick two points $A \equiv (x_1, y_1)$ and $B \equiv (x_2, y_2)$, and suppose $x_1 < x_2$ and $y_1 < y_2$. So we imagine this as two sides of a triangle:

     B
    /
   /
  /
 /
A

• The L1 norm is $|x_2 - x_1| + |y_2 - y_1|$. This is the distance on connecting to an origin $O$:

  δx
O----B
|   /
δy /
| /
|/
A

• The L2 norm is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, which is the distance of the vector $AB$, or the hypotenuse of the right angled triangle $AOB$:

  δx
O----B
|   /
δy / L2
| /
|/
A

• By triangle inequality, $OA + OB \geq AB$, hence $L_1 = \delta_x + \delta_y \geq L_2$