Proving the inequality $0\leq \frac{\sqrt{xy}}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} \leq 1$ Suppose $p\in(0,1)$. How might one show that 
\begin{equation}\tag{1}
0\leq \frac{\sqrt{xy}}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} \leq 1
\end{equation}
for all $x,y\in[0,1]$?
It is clearly non-negative, so the hard part is to show that it is never greater than 1. 
I was hoping to use a technique similar to the one to prove that 
$$
0\leq \sqrt{xy}\frac{\log x - \log y}{x-y}\leq 1
$$
for all $x,y\in[0,1]$. We can use an integral representation and see that 
\begin{align*}
\sqrt{xy}\frac{\log x - \log y}{x-y} 
&= \int_{0}^{\infty} \frac{\sqrt{xy}}{(x+t)(y+t)}dt\\
&\leq \int_{0}^{\infty} \frac{\sqrt{xy}}{(\sqrt{xy}+t)^2}dt\\
& = 1.
\end{align*}
Is there a suitable integral representation that can prove (1)?
 A: I've figured out the correct integral representation to use here. For $a\in(-1,1)$, consider the following integral representations:
\begin{align*}
\frac{x^a-y^a}{x-y} &= \frac{\sin(a\pi)}{\pi}\int_{0}^{\infty}\frac{t^a}{(x+t)(y+t)}dt\\
\text{and}\qquad ax^{a-1} &= \frac{\sin(a\pi)}{\pi}\int_{0}^{\infty}\frac{t^a}{(x+t)^2}dt.
\end{align*}
Similar to the example in the original post, we have
\begin{align*}
\frac{1}{a}\frac{x^a-y^a}{x-y} 
&\leq  \frac{\sin(a\pi)}{a\pi}\int_{0}^{\infty}\frac{t^a}{(\sqrt{xy}+t)^2}dt\\
 & = (\sqrt{xy})^{a-1}.
\end{align*}
Thus, if we let $a=1-p$, we have
\begin{align*}
\frac{1}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} 
 = \frac{1}{1-p}\frac{x^{\frac{1-p}{p}}-y^{\frac{1-p}{p}}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} 
&= \frac{1}{a}\frac{x^{\frac{a}{p}}-y^{\frac{a}{p}}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}}\\
&\leq \left(\sqrt{x^{\frac{1}{p}}y^{\frac{1}{p}}}\right)^{a-1} \\
& = \left(\sqrt{x^{\frac{1}{p}}y^{\frac{1}{p}}}\right)^{-p}\\
 &=\frac{1}{\sqrt{xy}} 
\end{align*}
which proves the desired result.
Hence, even though I only originally conjectured it for $p\in(0,1)$, the claim holds for $p\in(1,2)$ as well!
A: Here is what I have done so far.
This is incomplete,
but might help someone else.
The inequality is the same as
$\sqrt{xy}(x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}) 
\leq (1-p)(x^{\frac{1}{p}}-y^{\frac{1}{p}})
$
or,
letting
$\frac1{p} = q$,
$\sqrt{xy}(x^{q-1}-y^{q-1}) 
\leq (1-1/q)(x^{q}-y^{q})
$
where
$q > 1$.
Since
$\int t^{q-1}dt
=\frac{t^q}{q}
$,
or
$t^q
=q\int t^{q-1}dt
$,
this becomes
$\sqrt{xy}(x^{q-1}-y^{q-1}) 
\leq (1-1/q)q\int_y^x \int t^{q-1}dt
= (q-1)\int_y^x t^{q-1}dt
$.
Letting $r = q-1$,
this is
$\sqrt{xy}(x^r-y^r) 
\le r\int_y^x t^rdt
$
or
$\sqrt{xy}\frac{x^r-y^r}{x-y} 
\le \frac{r}{x-y}\int_y^x t^rdt
$
where
$r > 0$.
As a check,
letting $y \to x$,
this becomes
$x r x^{r-1}
\le r x^r
$
which is true.
For another check,
if $r=1$
this is
$\sqrt{xy}(x-y) 
\le \frac12(x^2-y^2)
$
or
$\sqrt{xy}
\le \frac12(x+y)
$
which is true.
I don't know where to go from here.
$t^r$ is not always
concave or convex
for different values of $r$,
so
perhaps considering
$r < 1$
and
$r > 1$
separately might work.
