Prove that $\frac{(\sum_{i=1}^m \|s_i - p\|_1)}{\sqrt2} \leq \sum_{i=1}^m \|s_i - p\|_2 \leq \sum_{i=1}^m \|s_i - p\|_1$ Prove that 
$$
\frac{1}{\sqrt2} \sum_{i=1}^m \|s_i - P\|_1
   \leq  \sum_{i=1}^m \|s_i - P\|_2
   \leq  \sum_{i=1}^m \|s_i - P\|_1
$$
Where $m$ is a number of points in 2d plane, e.g $s_i = (s_{x_i}, s_{y_i})$ and $p = (p_x, p_y)$
Note that
$$
\begin{split}
\|s_i - p\|_1 &= |s_{x_i} - p_x| + |s_{y_i} - p_y| \\
\|s_i - p\|_2 &= \sqrt{|s_{x_i} - p_x|^2 + |s_{y_i} - p_y|^2}
\end{split}
$$
$P$ is stationary point, $s_i$ is one of $m$ points around $P$
 A: It suffices to show that for $a = (a_1, a_2) \in \mathbb{R}^2$, $b = (b_1, b_2) \in \mathbb{R}^2$, it holds that
$$\frac{1}{\sqrt{2}}\|a - b\|_1 \leq \|a - b\|_2 \leq \|a - b\|_1.$$
This is quite straightforward to check by definition:
\begin{align*}
& \|a - b\|_2 \\
= & \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2} \\
\leq & \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2 + 2|a_1 - b_1||a_2 - b_2|} \\
= & \sqrt{\left(|a_1 - b_1| + |a_2 - b_2|\right)^2} \\
= & \|a - b\|_1.
\end{align*}
On the other hand, by AG inequality $xy \leq \frac{1}{2}(x^2 + y^2)$:
\begin{align*}
& \frac{1}{2}\|a - b\|_1^2 \\
= & \frac{1}{2}|a_1 - b_1|^2 + \frac{1}{2}|a_2 - b_2|^2 + |a_1 - b_1||a_2 - b_2| \\
\leq & \frac{1}{2}|a_1 - b_1|^2 + \frac{1}{2}|a_2 - b_2|^2 + \frac{1}{2}|a_1 - b_1|^2 + \frac{1}{2}|a_2 - b_2|^2 \\
= & (a_1 - b_1)^2 + (a_2 - b_2)^2 \\
= & \|a - b\|_2^2.
\end{align*}
A: Let us demostrate something else: $$\frac{ \|u\|_1}{\sqrt{2}} \leq   \|u\|_2 \leq   \|u\|_1$$ for every $u\in\mathbb{R}^2$.
Note that $\|u\|_1=|u_1|+|u_2|\ge\sqrt{|u_1|^2+|u_2|^2}=\|u\|_2$ (for the inequality, is enough with power both sides).
On the other hand, it is easy to show that for all $x,y\in\mathbb{R}$ we have $xy\le\frac{x^2+y^2}{2}$, which implies $\frac{x+y}{\sqrt{2}}\le \sqrt{x^2+y^2}$ wherever $x,y\ge 0$.Putting $x=|u_1|$ and $y=|u_2|$, we get the desired inequality.
Now, it is easy to conclude your inequality.
EDIT we have $2xy<x^2+y^2$. Then $x^2+2xy+y^2\le2x^2+2y^2$, which implies $\frac{(x+y)^2}{2}\le x^2+y^2$. Squaring both sides $\left|\frac{x+y}{\sqrt{2}}\right|\le\sqrt{x^2+y^2}$. Now, if $x,y\ge0$ the abs  symbol desapperars.
