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)?

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!
