# Solving inequation with 3 absolute values

• Feb 25th 2011, 11:00 PM
msokol89
Solving inequation with 3 absolute values
Hello,
$|x-1|+|x-2|-|x+3|>=x$

• Feb 25th 2011, 11:04 PM
Prove It
You might have to use the triangle inequality here...

$\displaystyle x \leq |x - 1| + |x - 2| - |x + 3|$

$\displaystyle x \leq |x| + |-1| + |x| + |-2| - (|x| + |3|)$, since $\displaystyle |a + b| \leq |a| + |b|$ by the Triangle Inequality

$\displaystyle x \leq |x| + 1 + |x| + 2 - |x| - 3$

$\displaystyle x \leq |x|$.

A number is only ever less than its absolute value if it's negative, so

$\displaystyle x \leq 0$.
• Feb 26th 2011, 12:25 AM
msokol89
Quote:

Originally Posted by Prove It
You might have to use the triangle inequality here...

$\displaystyle x \leq |x - 1| + |x - 2| - |x + 3|$

$\displaystyle x \leq |x| + |-1| + |x| + |-2| - (|x| + |3|)$, since $\displaystyle |a + b| \leq |a| + |b|$ by the Triangle Inequality

$\displaystyle x \leq |x| + 1 + |x| + 2 - |x| - 3$

$\displaystyle x \leq |x|$.

A number is only ever less than its absolute value if it's negative, so

$\displaystyle x \leq 0$.

I understand, thanks a lot

Edit:Unfortunately, I think there is a problem with solving this question with triangle inequation.
When you write -(|x|+|3|) instead of -|x+3| it ruins the inequation because you subtract bigger number.