How Does the Triangle Inequality Apply to Complex Fraction Inequalities?

[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 don't know other way to correctly type it.

Please explain your answers, and the reason you use such method to answer this question.

Thanks a lot for your much needed help.

 

[courtrigrad]

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 $$|\frac{x^{2}}{x^{2}+9}|\leq \frac{|x|+2}{9}$$

 

[0rthodontist]

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

Sorry about the display of the question, I really don't know other way to correctly type it.

There's a little trick called "parentheses."

 

[stokes]

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

 

[stokes]

Please guys need some advice... Thanks.

 

[chaoseverlasting]

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

 

[CRGreathouse]

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