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Let $\alpha\in (0,1/2)$ be a parameter. Is it true thet for every $x>y>0$ real numbers we have $$y^{-\alpha} - x^{-\alpha} \leq C y^{-\alpha -\frac{1}{2}} (x-y)^{1/2}$$ for some constant $C>0$ and every $\alpha\in (0,1/2)$?

I believe so, because I can't find any counterexample but I'm struggling to prove it as well.

Thanks!

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That is equivalent to: $$ \frac{1-(x/y)^{-\alpha}}{((x/y)-1)^{1/2}}\leq C\tag{1}$$ or to the statement: the function $f(z)=\frac{1-z^{-\alpha}}{\sqrt{z-1}}$ is bounded on $z\in(1,+\infty)$, or to the statement:

The function $$g(u) = \frac{1-(1+u^2)^{-\alpha}}{u}\tag{2}$$ is bounded on $\mathbb{R}^+$.

However, that is trivial since $g(u)$ is non-negative and continuous over $\mathbb{R}^+$, and: $$ \lim_{u\to 0^+}g(u)=\lim_{u\to +\infty}g(u)=0.\tag{3}$$

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    $\begingroup$ Oh! Thanks a lot! :) That was smart $\endgroup$ – Mark May 4 '15 at 18:09
  • $\begingroup$ @Mark: you're welcome. Please consider accepting the answer as soon as you can, if you find it useful. $\endgroup$ – Jack D'Aurizio May 4 '15 at 18:12

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