# Triangle inequality problem

1. Sep 16, 2006

### stokes

Hello all, I am having some difficulties with a question. Hope you guys can help shine some light on the situation.

Use the triangle Inequality and the fact that 0< |a| < |b| + => 1/|b| <
1/|a| to establish the following chains of inequalities.

|x-2 / x^2+9| "Is less than and equal to" |x| + 2 / 9

Sorry about the display of the question, I really dont know other way to correctly type it.

Thanks alot for your much needed help.

2. Sep 17, 2006

So the triangle inequality is: $$|x+y| \leq |x| + |y|$$. We are also given that $$0 < |x| < |y| \rightarrow \frac{1}{|y|} < \frac{1}{|x|}$$ and want to establish $$|\frac{x^{2}}{x^{2}+9}|\leq \frac{|x|+2}{9}$$

Last edited: Sep 17, 2006
3. Sep 17, 2006

### 0rthodontist

There's a little trick called "parentheses."

4. Sep 17, 2006

### stokes

Thanks for making the question much more understandable.... any help?

5. Sep 18, 2006

### stokes

6. Sep 19, 2006

### chaoseverlasting

is it |x-2| or |x^2| /{....} ?

7. Sep 19, 2006

### CRGreathouse

How big can $x^2/(x^2+9)$ get?