Could the inequality $0<x(1-y^{-\frac{1}{x}})<2,x, y \in \mathbb{R}^+$ be solved? I have tried a various of ways but seems impossible to solve. Is there a way to solve this inequality, or a way to proof there does not exist solution?

I need to find out the $x$ and $y$ such that the inequality satisfy.

Thanks very much~

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    $\begingroup$ This is not an inequality. Where is the $\geq$ or $\leq$ sign? And what is the other side of the expression? $\endgroup$ – Ragnar Feb 6 '14 at 9:18
  • $\begingroup$ And do you want to solve for $x$ of $y$? $\endgroup$ – Ragnar Feb 6 '14 at 9:24
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    $\begingroup$ What do you mean by solve? My solution $x=1,\;y=2$ still stands after you edit the question (both are strictly positive). $\endgroup$ – gammatester Feb 6 '14 at 9:36
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    $\begingroup$ A partial result: From $0<x(1-y^{-\frac{1}{x}})$ you get $1>y^{-\frac{1}{x}}=1/y^{\frac{1}{x}}$ or $y^{\frac{1}{x}}>1$, and therefore $y>1.$ $\endgroup$ – gammatester Feb 6 '14 at 9:48
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    $\begingroup$ @TooTone, I guess for $y\le e^2$ all $x>0$ are solutions, and for $y > e^2$ the value for $x$ with $x(1-y^{-\frac{1}{x}})=2$ could most probably expressed with the LambertW function (but Maple refuses to give a result for variable $y$) $\endgroup$ – gammatester Feb 6 '14 at 10:53

One have to separate in the study if three cases :

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The previous "sketch of the ranges of validity (green)" was not fully satisfising. It is remplaced by a more precise drawing. Thanks to some comments, the limit of $y$ when $x$ tends to infinity is $e²$, which is confirmed by asymptotic expansion.

  • $\begingroup$ That's excellent! Thank you so much!! $\endgroup$ – 1LiterTears Feb 8 '14 at 20:57

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