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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.

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I think you forgot the $\le 1$ on each of your implications ;D –  DanZimm May 22 '13 at 4:29
    
The lower bound can be calculated differently. Note that for $x,y,z$ at least $2$ must be on the same side of $0$, so (taking $x,y$ as this pair) $|x+y|=|x|+|y|$. Does this help? –  user45150 May 22 '13 at 4:30
    
Thanks User 45150 would you like to explain it in more detail, thanks –  juantheron May 22 '13 at 4:41
    
It is not good to use \mid for absolute value, since it give incorrect spacing. TeX.SE Absolute Value Symbols. –  Martin Sleziak Jun 10 '13 at 10:35

1 Answer 1

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.

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