Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can we prove that $\left|\frac{x^2y^3}{x^4+y^4}\right|\leq |x|+|y|$?

I got this as homework but don't even know where to start. I've tried developing $(x+y)^4$ but that didn't help to find a connection.

share|cite|improve this question
up vote 4 down vote accepted

If $x=0$ or $y=0$, then we got $0\leq 0$ else $|\frac{x^2y^3}{x^4+y^4}|\leq |\frac{x^2y^3}{2x^2y^2}|\leq \frac{1}{2}|y|\leq |x|+|y|$

share|cite|improve this answer
I actually have that solution separately. What I need to prove is that exact inequality somehow. – Grozav Alex Ioan Dec 4 '12 at 21:15
I don't see what you are looking for, this is a proof,isn't it ? (I'm sorry I'm french and my english is ugly :() – matovitch Dec 4 '12 at 21:22
It is a proof indeed. The professor gave us this alternative answer,involving $\frac{1}{2}y$, which he proved, and asked us to proof the other one as homework. – Grozav Alex Ioan Dec 4 '12 at 21:27
What "other one," @GrozavAlexIoan? You have one problem here. – Thomas Andrews Dec 4 '12 at 21:28
BTW, you need to handle specially the (obvious) cases of $x=0$ or $y=0$. – Thomas Andrews Dec 4 '12 at 21:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.