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It has been 2 days that i am trying to prove that the set E defined by $$E=\left\{\frac{xy}{x^{2} +3y} ;x >0;y >0\right\}$$ `doesn't have a upper bound,but without success I first supposed that E has a upper bound M and i tried to find x and y (depending on M) such that the expression obtained will be superior to M but this method seem a bit hazardous to me and i still don't have found the appropriate x and y. Thank you in advance!

Yagami
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    Along the path $y=x^2$ the function becomes $$f(x,x^2) = \frac{x}{4}$$ – Ninad Munshi Sep 09 '20 at 01:25
  • Ninad Munshi's suggestion is a good one and can be generalized: if you're trying to prove a multivariate function doesn't have some property, try and homogenize the degree. – Integrand Sep 09 '20 at 01:47

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Set $x=\sqrt{y}$. The expression then becomes

$$\frac{xy}{x^{2} +3y}=\frac{y^{3/2}}{y +3y}=\frac{y^{3/2}}{4y}=\frac{1}{4}\sqrt{y}$$

Since this is unbounded we are done.

QC_QAOA
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  • Please read here for information on formatting MathJax
    https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference     – QC_QAOA Sep 09 '20 at 02:04