# Prove $\left|\frac{x^2y^3}{x^4+y^4}\right|\leq |x|+|y|$

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.

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## 1 Answer

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

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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