A circle with a circumscribed and an inscribed square centered at the origin  of a rectangular coordinate system with positive  and  axes  and  is shown in each figure  to  below.

The inequalities
\[|x|+|y|\leq\sqrt{2(x^{2}+y^{2})}\leq 2\mbox{Max}(|x|, |y|)\]
are represented geometrically* by the figure numbered
$\textbf{(A)}\ I\qquad\textbf{(B)}\ II\qquad\textbf{(C)}\ III\qquad\textbf{(D)}\ IV\qquad\textbf{(E)}\ \mbox{none of these}$
* An inequality of the form $f(x, y) \leq g(x, y)$, for all $x$ and $y$ is represented geometrically by a figure showing the containment
$\{\mbox{The set of points }(x, y)\mbox{ such that }g(x, y) \leq a\} \subset\\ \{\mbox{The set of points }(x, y)\mbox{ such that }f(x, y) \leq a\}$
for a typical real number $a$.
First, note that the following inequality 
\[|x|+|y|\leq\sqrt{2(x^{2}+y^{2})}\leq 2\mbox{Max}(|x|, |y|)\]
represents the graphs $|x| + |y| = a$, $\sqrt{2(x^{2}+y^{2})} = a$, and $2\mbox{Max}(|x|, |y|) = a$. We don't actually have to worry about the inequality, we just want to find the picture that graphs those $3$ equations for some constant $a$. 
WLOG, we can set $a=1$. To make our life easier, we can also focus entirely on the first quadrant, since all $4$ graphs have different shapes in the first quadrant. Therefore, $x \geq 0$ and $y \geq 0$. 
First, we have $|x| + |y| = 1$. Since $x \geq 0$ and $y \geq 0$, this becomes $x + y = 1$. After a bit of rearranging, we get 
\[y = -x + 1\].

This eliminates $III$, which does not have a diagonal line in the first quadrant.
Next, we have $\sqrt{2(x^{2}+y^{2})} = 1$. Squaring both sides and dividing by $2$, we get 
\[x^2+y^2=\frac{\sqrt{2}}{2}^2\]
Note that this describes a circle with the center $(0,0)$ and radius $\frac{\sqrt{2}}{2}$. We can find that $(\frac{1}{2}, \frac{1}{2})$ is on the circle, which is also on the line $y = -x + 1$. Therefore, the circle intersects the line at that point. We can graph this as follows:

This eliminates $I$, which does not include the intersection point listed above.
Finally, we have $2\mbox{Max}(|x|, |y|) = 1$. Rearranging and substituting $x$ for $|x|$ and likewise for $y$, we get 
\[\mbox{Max}(x, y) = \frac{1}{2}\]
Note that this equation means that either $x \leq y = \frac{1}{2}$ or $y \leq x = \frac{1}{2}$. Therefore, from $x = 0$ to $x = \frac{1}{2}$, the graph looks like the line $y = \frac{1}{2}$. Then, from $y = 0$ to $y = \frac{1}{2}$, the graph looks like the line $x = \frac{1}{2}$. Graphing this, we get:

Thus, we can eliminate $IV$. However, at this point, we don't know if the answer is $II$ or none of the graphs, since this entire time we were only graphing in the first quadrant to simplify things. However, note that no matter what sign $x$ or $y$ is, the equations all surround the variables with absolute value signs or exponents and make the values positive. Therefore, the graphs are symmetric about the $x$ and $y$ axes. As a result, the final graph looks like this:

Therefore, we can choose $\boxed{\textbf{B}}$ as our answer.