Question based on Triangle Inequality $\displaystyle |x+y|\leq |x|+|y|$ If $x,y,z\in \mathbb{R}-\left\{0\right\}$. Then prove that  $\displaystyle 1\leq \frac{|x+y|}{|x|+|y|}+\frac{| y+z|}{| y |+| z |}+\frac{| z+x|}{| z |+| x |}\leq 3$ 
My Try:: Using Triangle Inequality
$\displaystyle | x+y | \leq | x |+| y |\Rightarrow \frac{| x+y|}{| x |+| y |}\le 1$
Similarly $\displaystyle | y+z | \leq | y |+| z |\Rightarrow \frac{| y+z|}{| y |+| z |}\le 1$
Similarly $\displaystyle | z+x | \leq | z |+| x | \Rightarrow \frac{| z+x|}{| z |+| x |}\le 1$
Now Add all three equations::
$\displaystyle \frac{| x+y|}{| x |+| y |}+\frac{| y+z|}{| y |+| z |}+\frac{| z+x|}{| z |+| x |}\leq 3$
But I did not Understand How can I calculate Lower value of the Given expression.
 A: Case 1: $x, y, z$ are all greater than $0$. 
Then the expression simplifies to 3 as we can get rid of the absolute signs.
Case 2: Two of $x, y, z$ are greater than $0$. 
Without loss of generality let $x, y$ be greater than $0$. Then the expression reduces to:
$$1 + \frac{|x+z|}{x+|z|} + \frac{|y+z|}{y+|z|}$$
The above expression reduces to $1$ when we set $x=y=-z$ and in every other situation the value of the expression is greater than $1$. Thus, $1$ is the minimum value that we have found so far.
Case 3: Only one of $x, y, z$ is greater than $0$.
Without loss of generality suppose that $z$ is greater than $0$. Then using the strategy outlined for case 4 we can flip this case to that of case 2 an show that the minimum is again 1.
Case 4: $x, y, z$ are all less than $0$. 
Just as in case 1, the expression simplifies to 3. We can see this as follows:
Define X=-x, Y=-y and Z=-z. By definiton, $X, Y, Z$ are all greater than 0. Thus,
$$\frac{\mid x+y\mid}{\mid x \mid+\mid y \mid}+\frac{\mid y+z\mid}{\mid y \mid+\mid z \mid}+\frac{\mid z+x\mid}{\mid z \mid+\mid x \mid}$$
is equivalent to:
$$\frac{\mid -X+-Y\mid}{\mid -X \mid+\mid -Y \mid}+\frac{\mid -Y+-Z\mid}{\mid -Y \mid+\mid -Z \mid}+\frac{\mid -Z+-X\mid}{\mid -Z \mid+\mid -X \mid}$$
Factor out the '-1's and we are back to case 1.
