Does this hold for these absolute values 
Suppose $\left|x_{1}\right|\ge\left|x_{2}\right|$, $\left|y_{1}\right|\ge\left|y_{2}\right|$
  and $\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\left(x_{1}-y_{2}\right)\left(x_{2}-y_{1}\right)<0$,
Can anyone show or is it true that $\left|x_{1}+x_{2}-y_{1}-y_{2}\right|\ge\left|\left|x_{1}\right|-\left|y_{1}\right|\right|+\left|\left|x_{2}\right|-\left|y_{2}\right|\right|$
  ?

 A: By replacing $(x_1,x_2,y_1,y_2)\leftrightarrow (-x_1,-x_2,-y_1,-y_2)$ we see that the hypothesized conditions all still hold when we change all numbers to their negatives. 
By replacing $(x_1,x_2, y_1,y_2) \leftrightarrow (y_1,y_2,x_1,x_2)$ we see that the hypothesized conditions all still hold when we swap $x$ and $y$. 
So without loss of generality we can assume that $x_1 \geq 0$ and $|x_1| > |y_1|$. 
This requires then 
$$ (x_2 - y_2)(x_1 - y_2)(x_2 - y_1) < 0 \tag{*} $$
We consider separately the cases where $x_2 - y_2 < 0$ or $> 0$.
Case 1: $x_2 - y_2 > 0$. 
Then
$$ |x_1 - y_1 + x_2 - y_2| = |x_1 - y_1| + |x_2 - y_2| \geq ||x_1| - |y_1|| + ||x_2| - |y_2|| $$ 
by triangle inequality:
$$ |a| \leq |a-b| + |b| \implies |a| - |b| \leq |a-b| $$
and
$$ |b| \leq |b-a| + |a| \implies |b| - |a| \leq |b-a| $$
Case 2: $x_2 - y_2 < 0$. 
By (*) it is necessary that $(x_1 - y_2)(x_2 - y_1) > 0$. So either the two terms are both negative or both positive. The first option is not possible by assumption that $x_1 > |y_1| \geq |y_2|$. So we have that $x_2 - y_1 > 0$. Combining with $x_2 - y_2 < 0$ this implies that $y_1 < 0$. So it suffices to show that 
$$  x_1 + x_2  + |y_1| - y_2 \geq x_1 - |y_1| + ||x_2| - |y_2||  \tag{**}$$
Now either $|x_2| > |y_2|$ in which case $x_2 > 0$ or $|x_2| < |y_2|$. In the first case it suffices to show that
$$  x_1 + x_2 + |y_1| - y_2 \geq x_1 - |y_1| + x_2 - |y_2| $$
which follows from the assumption $|y_1| \geq |y_2|$. In the second case it suffices to show, cancelling the $x_1$, 
$$ x_2 + |y_1| - y_2 \geq -|y_1| + |y_2| - |x_2| $$
but the left hand side is strictly positive while the right hand side is non-positive as $|y_1| \geq |y_2|$. This concludes the proof. 
