inequality with absolute values
Recalling that the absolute value^{} (http://planetmath.org/AbsoluteValue) of a real number means on the number line (real axis^{}) the distance of the point from the origin, we have the following three rules which remove the absolute value signs from an inequality (we use the logical symbol “$\vee $” for alternativeness ‘or’). Note that the symbols “$\ge $” and “$\le $” may also be without the equality bar.

1.
$a\ge b\mathit{\hspace{1em}\hspace{0.25em}}\iff \mathit{\hspace{1em}}a\le b\vee a\ge b$

2.
$a\le b\mathit{\hspace{1em}\hspace{0.25em}}\iff \mathit{\hspace{1em}}b\le a\le b$

3.
$a\ge b\mathit{\hspace{1em}}\iff \mathit{\hspace{1em}}{a}^{2}\ge {b}^{2}$
These rules are valid for all real values of $a$ and $b$. For example, if one has a case
$$ 
corresponding the rule 2, this inequality seems to be impossible since no absolute value is negative; but now also the result $$ given by the rule 2 is impossible — no real number is simultaneously greater than $+5$ and less than $5$.
Examples. We solve some inequalities with absolute values.
a) $2x+1>5x$
$$ or $2x+1>5x$ (rule 1)
$$ or $3x>1$
$$ or $$
$$ (combined)
b) $8x+x2>6$
$8x>6x2$
$$ or $8x>6x2$ (rule 1)
$x2>8x+6$ or $x2>68x$
$$ or $x2>8x+6$ or $$ or $x2>68x$ (rule 1 twice)
$$ or $7x>8$ or $$ or $9x>8$
$$ or $$ or $x>4/7$ or $x>8/9$
$$ or $x>4/7$ (from the number line)
c) $15x\le 3$
$3\le 15x\le 3$ (rule 2)
$4\le 5x\le 2$ (subtracted 1 from all parts)
$4/5\ge x\ge 2/5$ (divided by $5)$
$2/5\le x\le 4/5$ (rewritten from end to begin)
Title  inequality with absolute values 

Canonical name  InequalityWithAbsoluteValues 
Date of creation  20130322 16:57:20 
Last modified on  20130322 16:57:20 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  7 
Author  pahio (2872) 
Entry type  Topic 
Classification  msc 97D40 
Related topic  AbsoluteValue 
Related topic  AbsoluteValueInequalities 
Related topic  OrderOfSixMeans 