Could the inequality $0

Question

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~

This is not an inequality. Where is the $\geq$ or $\leq$ sign? And what is the other side of the expression? — Ragnar, Feb 06 '14 at 09:18
$x=$ whatever, $y=1$ leads to zero and the left inequality is not satisfied. Did you miss typing the possibility of equalities ? — Claude Leibovici, Feb 06 '14 at 09:33
What do you mean by solve? My solution $x=1,;y=2$ still stands after you edit the question (both are strictly positive). — gammatester, Feb 06 '14 at 09:36
@gammatester You're right solve isn't clear. But it might be along the lines of if you were asked to solve $y<x$ you would say it's the area under the line $y=x$. — TooTone, Feb 06 '14 at 09:42
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 $y>1.$ — gammatester, Feb 06 '14 at 09:48
Another partial result $y=\left(\frac{x}{x-a}\right)^x$ is the line which upperbounds the expression to $a$. — Claude Leibovici, Feb 06 '14 at 09:54
@gammatester, building on your constraint $y>1$, and example $(1,2)$ an obvious set of points that satisfy the inequality is ${(x,y)}$ such that $x\le2,y>1$. — TooTone, Feb 06 '14 at 10:29
@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 $y$) — gammatester, Feb 06 '14 at 10:53
@gammatester Cool, I plotted the values numerically and could see $y$ tending to a value around $7\approx e^2$ as $x\to\infty$. — TooTone, Feb 06 '14 at 11:03

JJacquelin · Accepted Answer · 2014-02-06T13:44:11.440

3

One have to separate in the study if three cases :

enter image description here

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.

edited Feb 06 '14 at 13:44

answered Feb 06 '14 at 10:57

JJacquelin

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That's excellent! Thank you so much!! – 1LiterTears Feb 08 '14 at 20:57

Could the inequality $0

1 Answers1