Inequality on a root Does it hold that 
$$
\left|\max\left(\sqrt{\left|z\right|},\sqrt{\left|x\right|}\right)-\max\left(1,\sqrt{\left|y\right|}\right)\right|+\left|\min\left(\sqrt{\left|z\right|},\sqrt{\left|x\right|}\right)-\min\left(1,\sqrt{\left|y\right|}\right)\right|\le\sqrt{z-y+x+1}
$$
for all $x$, $y$, $z$ and $w$ with $-1<x<y<z$?
 A: Disclaimer: Lots of cases to consider, as expected when you see all the max, min and absolute values. Main idea is to smooth towards the case where $x, y, z, w$ are non-negative, then proceed to prove that case. I replace $x<y<z<w$ with the weaker condition $x \leq y \leq z \leq w$.
For real numbers $x \leq y \leq z \leq w$, define 
$$f(x, y, z, w)=\sqrt{w-z+y-x}$$
\begin{align}
g(x, y, z, w)&=\left|\max\left(\sqrt{\left|w\right|},\sqrt{\left|y\right|}\right)-\max\left(\sqrt{\left|x\right|},\sqrt{\left|z\right|}\right)\right| \\
& +\left|\min\left(\sqrt{\left|w\right|},\sqrt{\left|y\right|}\right)-\min\left(\sqrt{\left|x\right|},\sqrt{\left|z\right|}\right)\right|
\end{align}
This is so that I don't have to keep writing those expressions. The inequality becomes $f(x, y, z, w) \geq g(x, y, z, w)$.
It is clear that $-w\leq -z \leq -y \leq -x$, $f(x, y, z, w)=f(-w, -z, -y, -x)$ and $g(x, y, z, w)=g(x, y, z, w)$.
Thus if $z<0$, then $-y \geq -z>0$ and it suffices to prove $f(-w, -z, -y, -x) \geq g(-w, -z, -y, -x)$.
It thus suffices to consider the case where $w \geq z \geq 0$. 
This is where we start considering many cases. The idea is this:
Lemma: If $x<0$ and $w \geq z \geq 0$, then if we consider $a \leq b \leq c \leq d$ such that $\{a, b, c, d\}=\{|x|, |y|, z, w\}$, then $f(x, y, z, w) \geq f(a, b, c, d)$ and $g(x, y, z, w)=g(a, b, c, d)$
Before I prove this lemma, I will explain its applications to the problem, before we get too bogged down with the 10 cases we consider in the proof of the lemma.
Let's suppose for now we have proven the lemma. Suppose $x<0$ and $w \geq z \geq 0$. Note that to prove $f(x, y, z, w) \geq g(x, y, z, w)$, it suffices to prove $f(a, b, c, d) \geq g(a, b, c, d)$. Note that $\{a, b, c, d\}=\{|x|, |y|, z, w\}$, so $a, b, c, d$ are all non-negative. It thus suffices to prove the case where $0 \leq x \leq y \leq z \leq w$.
Let us now proceed to prove the lemma:
Case 1: $y \geq 0$. We have 4 subcases.
Case 1a) $-y \leq x<0 \leq y \leq z \leq w$.
Then $0< -x \leq y \leq z \leq w$.
$g(x, y, z, w)= (\sqrt{w}-\sqrt{z})+(\sqrt{y}-\sqrt{-x})=g(-x, y, z, w)=g(a, b, c, d)$
\begin{align}
f(x, y, z, w) \geq f(a, b, c, d)=f(-x, y, z, w) & \Leftrightarrow \sqrt{w-z+y-x} \geq \sqrt{w-z+y-(-x)} \\
& \Leftrightarrow w-z+y-x \geq w-z+y+x \\
& \Leftrightarrow 0 \geq 2x
\end{align}
which is true.
Case 1b) $-z \leq x<-y \leq 0 \leq y \leq z \leq w$
Then $0 \leq y<-x \leq z \leq w$
$g(x, y, z, w)=(\sqrt{w}-\sqrt{z})+(\sqrt{-x}-\sqrt{y})=g(y, -x, z, w)=g(a, b, c, d)$
\begin{align}
f(x, y, z, w) \geq f(a, b, c, d)=f(y, -x, z, w) & \Leftrightarrow \sqrt{w-z+y-x} \geq \sqrt{w-z+(-x)-y} \\
& \Leftrightarrow w-z+y-x \geq w-z-x-y \\
& \Leftrightarrow 2y \geq 0
\end{align}
which is true.
Case 1c) $-w \leq x<-z \leq 0 \leq y \leq z \leq w$
Then $0 \leq y \leq z<-x \leq w$.
$g(x, y, z, w)=(\sqrt{w}-\sqrt{-x})+(\sqrt{z}-\sqrt{y})=g(y, z, -x, w)=g(a, b, c, d)$
\begin{align}
f(x, y, z, w) \geq f(a, b, c, d)=f(y, z, -x, w) & \Leftrightarrow \sqrt{w-z+y-x} \geq \sqrt{w-(-x)+z-y} \\
& \Leftrightarrow w-z+y-x \geq w+x+z-y \\
& \Leftrightarrow 2y \geq 2(x+z)
\end{align}
which is true.
Case 1d)$x<-w \leq 0 \leq y \leq z \leq w$
Then $0 \leq y \leq z \leq w<-x$.
$g(x, y, z, w)=(\sqrt{-x}-\sqrt{w})+(\sqrt{z}-\sqrt{y})=g(y, z, w, -x)=g(a, b, c, d)$
\begin{align}
f(x, y, z, w) \geq f(a, b, c, d)=f(y, z, w, -x) & \Leftrightarrow \sqrt{w-z+y-x} \geq \sqrt{(-x)-w+z-y} \\
& \Leftrightarrow w-z+y-x \geq -x-w+z-y \\
& \Leftrightarrow 2(w+y) \geq 2z
\end{align}
which is true.
Case 2: $y<0$. We have 6 subcases.
Case 2a) $-z\leq x \leq y<0 \leq z \leq w$
Then $0<-y \leq -x \leq z \leq w$
$g(x, y, z, w)=(\sqrt{w}-\sqrt{z})+(\sqrt{-x}-\sqrt{-y})=g(-y, -x, z, w)=g(a, b, c, d)$
\begin{align}
f(x, y, z, w) \geq f(a, b, c, d)=f(-y, -x, z, w) & \Leftrightarrow \sqrt{w-z+y-x} \geq \sqrt{w-z+(-x)-(-y)} \\
\end{align}
which is true.
Case 2b) $-w \leq x<-z \leq y<0 \leq z \leq w$
Then $0<-y \leq z<-x \leq w$.
$g(x, y, z, w)=(\sqrt{w}-\sqrt{-x})+(\sqrt{z}-\sqrt{-y})=g(-y, z, -x, w)=g(a, b, c, d)$
\begin{align}
f(x, y, z, w) \geq f(a, b, c, d)=f(-y, z, -x, w) & \Leftrightarrow \sqrt{w-z+y-x} \geq \sqrt{w-(-x)+z-(-y)} \\
& \Leftrightarrow w-z+y-x \geq w+x+z+y \\
& \Leftrightarrow 0 \geq 2(x+z)
\end{align}
which is true.
Case 2c) $x<-w \leq -z \leq y<0 \leq z \leq w$
Then $0 \leq -y \leq z \leq w<-x$.
$g(x, y, z, w)=(\sqrt{-x}-\sqrt{w})+(\sqrt{z}-\sqrt{-y})=g(-y, z, w, -x)=g(a, b, c, d)$
\begin{align}
f(x, y, z, w) \geq f(a, b, c, d)=f(-y, z, w, -x) & \Leftrightarrow \sqrt{w-z+y-x} \geq \sqrt{(-x)-w+z-(-y)} \\
& \Leftrightarrow w-z+y-x \geq -x-w+z+y \\
& \Leftrightarrow 2w \geq 2z
\end{align}
which is true.
Case 2d) $-w \leq x \leq y<-z \leq 0 \leq z \leq w$
Then $0 \leq z<-y \leq -x \leq w$.
$g(x, y, z, w)=(\sqrt{w}-\sqrt{-x})+(\sqrt{-y}-\sqrt{z})=g(z, -y, -x, w)=g(a, b, c, d)$
\begin{align}
f(x, y, z, w) \geq f(a, b, c, d)=f(z, -y, -x, w) & \Leftrightarrow \sqrt{w-z+y-x} \geq \sqrt{w-(-x)+(-y)-z} \\
& \Leftrightarrow w-z+y-x \geq w+x-y-z \\
& \Leftrightarrow 2y \geq 2x
\end{align}
which is true.
Case 2e) $x<-w \leq y<-z \leq 0 \leq z \leq w$
Then $0 \leq z<-y \leq w<-x$.
$g(x, y, z, w)=(\sqrt{-x}-\sqrt{w})+(\sqrt{-y}-\sqrt{z})=g(z, -y, w, -x)=g(a, b, c, d)$
\begin{align}
f(x, y, z, w) \geq f(a, b, c, d)=f(z, -y, w, -x) & \Leftrightarrow \sqrt{w-z+y-x} \geq \sqrt{(-x)-w+(-y)-z} \\
& \Leftrightarrow w-z+y-x \geq -x-w-y-z \\
& \Leftrightarrow 2(w+y) \geq 0
\end{align}
which is true.
Case 2f) $x \leq y<-w \leq -z \leq 0 \leq z \leq w$
Then $0 \leq z \leq w<-y \leq -x$.
$g(x, y, z, w)=(\sqrt{-x}-\sqrt{-y})+(\sqrt{w}-\sqrt{z})=g(z, w, -y, -x)=g(a, b, c, d)$
\begin{align}
f(x, y, z, w) \geq f(a, b, c, d)=f(z, w, -y, -x) & \Leftrightarrow \sqrt{w-z+y-x} \geq \sqrt{(-x)-(-y)+w-z} \\
& \Leftrightarrow w-z+y-x \geq -x+y+w-z \\
\end{align}
which is true.
Lemma proved.
Finally, we proceed to consider the case where $0 \leq x \leq y \leq z \leq w$.
The inequality becomes $\sqrt{w-z+y-x} \geq \sqrt{w} - \sqrt{z} + \sqrt{y} - \sqrt{x}$.
Note that 
\begin{align}
\sqrt{w-z+y-x} + \sqrt{z} \geq \sqrt{w}+ \sqrt{y-x} & \Leftrightarrow (\sqrt{w-z+y-x} + \sqrt{z})^2 \geq (\sqrt{w}+ \sqrt{y-x})^2 \\
& \Leftrightarrow (w-z+y-x)z \geq w(y-x) \\
& \Leftrightarrow (w-z)(z+x-y) \geq 0 
\end{align}
and 
\begin{align}
\sqrt{y-x} + \sqrt{x} \geq \sqrt{y} & \Leftrightarrow (\sqrt{y-x} + \sqrt{x})^2 \geq y \\
& \Leftrightarrow x(y-x) \geq 0
\end{align}
Summing gives the desired inequality, so we are done. (Finally...) 
That took forever to type. If anyone has an elegant proof with less cases, that would be great. The proof for the various cases are almost identical, so there might be a way to cover all cases at one go.
A: If $x, y, z, w = 1, 4, 9, 16$,
this becomes
$\sqrt{16-9+4-1} \le |\max(4,2)-\max(1,3)|+|\min(4,2)-\min(1,3)|$
or
$\sqrt{10} \le |4-3|+|2-1| = 1 + 1 = 2$
which seems to be false.
Note: This was the simplest example I could come up with.
Also, from the inequalities,
you can rewrite this as,
letting $x, y, z, w = a^2, b^2, c^2, d^2$
with $0 < a < b < c < d$,
$\sqrt{d^2-c^2+b^2-a^2} \le |d-c|+|b-a|$,
which looks a lot easier to analyze.
A: If they are all positive, $x \lt y \lt z \lt w$ tells us what the max and min are, so you are asking whether $\sqrt {w-z+y-x} \ge \sqrt w -\sqrt z+\sqrt y - \sqrt x$.  The two element version is true:  you are asking whether $\sqrt {y-x} \ge \sqrt y - \sqrt x$  You can divide through by $\sqrt x$ to ask whether $\sqrt {\frac yx -1}-\sqrt {\frac yx}+1 \gt 0$ with $\frac yx \gt 1$.  You can prove this by taking a derivative or just by looking at the graph.  The four element version appears to be correct, but I haven't found a proof.
