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)?

  • $\begingroup$ You must have a typo in (1), because the numerator is zero. Did you mean $x^{1/p-1}-y^{1/p-1}$? Also the title doesn't format correctly. $\endgroup$ – Χpẘ Apr 14 '17 at 16:53

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!

  • $\begingroup$ Also, the example with the logarithm is recovered in the limit $p\rightarrow 1$. $\endgroup$ – luftbahnfahrer Apr 21 '17 at 18:00

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.


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