Let x,y,z be non-zero reals

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Let x,y,z be non-zero reals

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#1 h Jul 16, 2019, 2:23 PM • 1 Y

Y by Adventure10

Let $x,y,z$ be non-zero real numbers. Define $E=\frac{|x+y|}{|x|+|y|}+\frac{|x+z|}{|x|+|z|}+\frac{|y+z|}{|y|+|z|}$ , then the number of all integers which lies in the range of $E$ is equal to.

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#2 h Jul 16, 2019, 2:42 PM • 3 Y

Y by Adventure10, Mango247, Mango247

I think that $E\le 3$ it is evident if $x,y,z>0$ or $x,y,z<0$

This post has been edited 1 time. Last edited by TuZo, Jul 16, 2019, 2:43 PM

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#3 h Jul 16, 2019, 2:45 PM • 1 Y

Y by Adventure10

Clearly $E$ is a non-negative quantity. Also by triangle inequality, we can easily see that $E\le 3.$ Hence $0\le E\le 3.$ But
$E=0\implies |x+y|=|y+z|=|z+x|=0$ $\implies x+y=y+z=z+x=0\implies x=y=z=0$ which contradicts the fact that $x,y,z$ are non-zero real numbers. Hence we conclude that $0<E\le 3.$ So the set of all integers lying in the range of $E$ is a subset of $\{1,2,3\}.$

$x=1,y=-1,z=1\implies E=1.$
$x=1,y=1,z=-3\implies E=2.$
$x=1,y=1,z=1\implies E=3.$

So $\{1,2,3\}$ is the required set.

This post has been edited 1 time. Last edited by Bashy99, Jul 16, 2019, 2:46 PM

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#4 h Apr 16, 2025, 1:08 PM

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Let $x,y,z$ be real numbers. Prove that $\frac{|x-y|}{1+|x|+|y|}\leq\frac{|x-z|}{1+|x|+|z|}+\frac{|z-y|}{1+|z|+|y|}$ Let $x,y,z$ be real numbers. Prove that
$\frac{|x-y|}{1+|x-y|} \leq \frac{|x-z|}{1+|x-z|} + \frac{|z-y|}{1+|z-y|}$ h

This post has been edited 1 time. Last edited by sqing, Apr 16, 2025, 1:10 PM

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