# Is this inequality of real numbers true?

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!

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: $$\lim_{u\to 0^+}g(u)=\lim_{u\to +\infty}g(u)=0.\tag{3}$$