How find all values of $\frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}$ 
$$A =
 \frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}$$
Where $x, y $ and $z$ are real numbers and non-zero. How can we find
  all values of $A$?

My try:
From $x, y$ and $z$ at least two of have same sign $\longrightarrow A \geq 1$ 
$$\left\{
\begin{array}{}
|x+y| \leq |x| + |y| \longrightarrow \frac{|x+y|}{|x|+|y|} \leq 1 \\
|z+y| \leq |z| + |y| \longrightarrow \frac{|z+y|}{|z|+|y|} \leq 1 \\
|x+z| \leq |x| + |z| \longrightarrow \frac{|x+z|}{|x|+|z|} \leq 1 \\
\end{array}
\right. \longrightarrow A \leq 3$$
Also I have an example for $A = 3$ and $A = 1$:
$A=3 \longleftarrow x=y=z\neq0$
$A=1 \longleftarrow x=y\neq0$ and $z=-x$
But still I can't find all values of $A$.
 A: For $t\in[0,1]$ let $(x,y,z)=(1,1,-t)$. Then $$A(x,y,z)=1+2\frac{1-t}{1+t}.$$ When $t=0$ this is $3$, when $t=1$, this is zero. Since $A(x,y,z)$ is continuous on $\mathbb R^3\setminus\{(0,0,0)\}$, $A$ must take all values between $1$ and $3$, by the intermediate value theorem. Or we can find $t$ based on $A$:
$$\frac{A+1}{2}=\frac{2}{1+t}$$ or $t=\frac{4}{A+1}-1.$ Show when $A\in [1,3]$ then $t\in[0,1]$.
Now you need to prove that $1\leq A\leq 3$ in general. $A\leq 3$ is easy, since each term is $\leq 1$ by the triangle inequality. $A\geq 1$ is because one pair of $x,y$ must be the same sign (or one zero) in which case one of the terms is always $1$.
A: You basically already solved this, you just need to think of this from a different PoV:
Think of A as a function $A(x,y,z)=\frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}.$
Now because you divide by sum of 2 positive numbers this function exist for all values, in other words continuous. 
Because you know that A can't be less than 1 and A can't be more than 3 and A is continuous it has to go from 1 to 3 somehow, all the ways to do that have something in common, they all pass through all the numbers between 1 and 3 because if they don't it means there is a hole in the function, but we already know that the function is continuous, therefore $1\le A\le 3$
