About the inequality $rx^{r-1}\left(\sqrt{xy}-y\right)>x^{r}-y^{r}$ for $x>y>0$ and $r<0$

Question

In my book (Dictionnary of Inequalities ed 2 by Peter Bullen) we have p:27

If $x>y>0$ and $r<0$ then :

$$rx^{r-1}\left(x-y\right)>x^{r}-y^{r}>ry^{r-1}\left(x-y\right)$$

The above inequality is easy dividing by $x-y$ and using Hermite-Hadamard inequality .

Now I ask for a refinement of it :

If $x>y>0$ and $r<0$ then :

$$rx^{r-1}\left(\sqrt{xy}-y\right)>x^{r}-y^{r}\tag{I}$$

I have two question :

How to show $(I)$ ?

What is the limit for $x\to \infty$ (RHS-LHS in $(I)$)?

Thanks for all your reply and your effort in this sense .

Answer 1

This seems to be true

Considering the function $$f(x,y)=rx^{r-1}\left(\sqrt{xy}-y\right)-(x^{r}-y^{r})$$

Part 1

Let $y=k x$ with $0 < k <1$.

$$f(x,k x)=g(x)=\left(k^r-k r+\sqrt{k} r-1\right) x^r$$ So, to have $g(x)>0$ for any value of $k$ and $r<0$ it remains to show that $$h(r)=k^r-k r+\sqrt{k} r-1 >0$$ $$h'(r)=k^r \log (k)-k+\sqrt{k}\qquad \text{and} \qquad h''(r)=k^r \log ^2(k) ~~> ~~0$$ $$h'(r)=0 \implies r_*=\frac{1}{\log (k)}\log \left(-\frac{\sqrt{k}-k}{\log (k)}\right) > 0$$ So, $h(r)$ is a decreasing function which is always positive and then $f(x,kx) >0$.

Part 2

If, as commented by @Martin R, we make $y=1$, we have $$f(x,1)=g(x)=1-x^{r-1} \left(x-r \sqrt{x}+r\right)$$ $$g'(x)=\frac{1}{2} r \Big[(2 r-1) \sqrt{x}-2 x+2-2r\Big] x^{r-2}$$

$g'(x)$ cancels for two values of $x$ $$x_\pm=\frac{1}{8} \left(4 r^2-12 r+9 \pm(2 r-1) \sqrt{4 r^2-20 r+17}\right)$$ but $x_+ <1$.

But $g(x_-) >0$ as well as $g''(x_-)$.