# Inequality problem

1. Mar 21, 2004

### exequor

How do i show that the inequality
|x-3| >= |x|-|3| holds for any x contained in R (real values).

and in the function f(x) = (x^2)/(1-x^2) would it be correct to say that x^2 cannot be equal to 1 because that would cause the function to have no real solution?

2. Mar 21, 2004

### Chen

Raise both sides of the 'equation' to the power of 2:
$$(x - 3)^2 \geq x^2 - 2|x||3| + 9$$
$$x^2 - 6x + 9 \geq x^2 - 2|x||3| + 9$$
$$-6x \geq -6|x|$$
$$x \leq |x|$$
True for every real number.

You can say that the function f(x) is not defined for $$x^2 = 1$$. x can have whatever value you, but the function won't be defined at these points.

3. Mar 21, 2004

### matt grime

Altenatively,

|x-3| = x-3

if x>=3

when |x|-3 = x-3 also, so it's true for x>=3,

if you consider 3>x>0 , and 0>x you can work out those cases.

Of course, neither Chen's nor my proof works for complex numbers. You might want to look up things like the triangle inequality to see why

| |x|-|y|| <= |x-y| <= |x|+|y|

for complex numbers too.

4. Mar 21, 2004

### Hurkyl

Staff Emeritus
*cringe*

Actually, Chen made a common mistake. Starting from $|x-3| \geq |x| - 3$, he correctly deduced that $x \leq |x|$. However the steps don't work in the reverse order. (In particular, "raising both sides to the power of 2" isn't a reversible operation)

You can kinda reverse it via the deduction:

if $a^2 \geq b^2$ then $|a| \geq |b|$. (which you may or may not need to prove before your teacher lets you use it)

In general, though, the most straightforward way to solve this type of problem is to split the domain into several cases. Since you have $|x|$ and $|x-3|$ in your problem, you want to split the domain into three cases, one for each combination of $x$ and $x-3$ being positive or negative.

It would be correct to say:

If x is in the domain of f, then x2 cannot be equal to 1.

5. Mar 21, 2004

### uart

Or in other words, inequalities generally do not withstand non-monotonic operations on both sides. With monotonic operations the inequality remains intact (or the direction of the inequality simply reverses for the case of monotone decreasing), but non-monitonic functions "bust up" the inequality.

6. Mar 22, 2004

### arildno

The simplest way of doing this is to prove the inequality |x|<=|3|+|x-3|, which follows from the standard triangle inequality