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$$A = \frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}$$

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 $A$.

Amiri
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  • apply $|x|+|y|<|x|+|y|+|z|$ in three items. – Nosrati Aug 17 '17 at 15:32
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    You can get all values between $1$ and $3$ by taking any path between $(1,1,1)$ and $(1,1,-1)$ that does not pass through $(0,0,0)$, like $\gamma(t)=(1,1,t)$ for $t\in[-1,1]$. Since your expression is continuous, the path must take the value to all values in $(1,3)$. Now show that it is not possible for $A>3$ or $A<1$. – Thomas Andrews Aug 17 '17 at 15:39

2 Answers2

3

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 $1$.

Thomas Andrews
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1

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 $1\le A\le 3$

amWhy
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ℋolo
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  • Actually I don't know about continuous functions – Amiri Aug 17 '17 at 16:03
  • @Esmaeeli continuous function is a function that exist to all values, because your function is only adding and division by non-zero to all A values between the minimum(you showed it is 1) and the maximum(you showed it is 3) there has to be x,y and z that will give that result. btw about your question of what R^3∖{(0,0,0)} means, it means to all values of R^3 Apart from (0,0,0) – ℋolo Aug 17 '17 at 16:06