2
$\begingroup$

For every $x,y,z \in \mathbb{R} $

$$ |x|+|y|+|z| \leq |x+y-z|+|x-y+z|+|-x+y+z| $$ I tried to prove it by cases, but I have troubles when the three numbers have the same sign.

2 Answers 2

2

Because $$\sum_{cyc}|x+y-z|=\frac{1}{2}\sum_{cyc}\left(|x+y-z|+|x+z-y|\right)\geq\frac{1}{2}\sum_{cyc}|x+y-z+x+z-y|=\sum_{cyc}|x|$$

1

if all of the numbers are positive then the left is $x+y+z$ and the right is at least $x+y+z$ (that is what you get when you remove the absolute values).

Notice that if all three are negative all of the expressions have the same values as with $(-x,-y,-z)$.

  • 0
    And I suppose checking all of the in between $(\pm x,\pm y,\pm z)$ result similarly. – 2017-01-24
  • 0
    well, he said he was having trouble with the case in which they all had the same sign. – 2017-01-24
  • 0
    The right side may be less than that. Consider the case when the three numbers are equal to 1, the right side is not 6, but 3. – 2017-01-24
  • 0
    @JJmz fixed, thanks. – 2017-01-24