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I think the following two inequalities are true. However, the proof may not be easy. Does anyone have any hints? Thank you very much!

Fix $a>1$. there exists two constants $K_1$ and $K_2$, such that

$$ \frac{1}{1+|x-y|^{1+a}}\frac{1}{1+|x-z|^{1+a}} \le \frac{K_1}{1+|x-(y+z)/2|^{1+a}} \le \frac{K_2}{1+|x-y|^{(1+a)/2}}\frac{1}{1+|x-z|^{(1+a)/2}} $$

holds for all $x,y,z\in R$.

Thanks a lot!

Anand

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Where did these arise? –  Alex Becker Feb 24 '12 at 9:31
    
In my research, I would like to have this inequality. These inequalities are my guess. For sure, I can modify accordingly. It is related to the Poisson kernel and fractional heat equations. –  Anand Feb 24 '12 at 11:51
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1 Answer

up vote 1 down vote accepted

Let $\alpha=|x-y|$, $\beta=|x-z|$, $\gamma=|x-(y+z)/2|$ and $p=1+a>2$. It is clear that $\gamma\le(\alpha+\beta)/2\le\max(\alpha,\beta)$. Assume without loss of generality that $\alpha\le\beta$. Then $$ \frac1{1+\alpha^p}\,\frac1{1+\beta^p}\le\frac1{1+\beta^p}\le\frac1{1+\gamma^p}, $$ proving the first inequality with constant $K_1=1$. This constant is sharp, as can be seen by letting $\alpha=\beta=\gamma\to0$.

The second inequality reads now as $$ \frac1{1+\gamma^p}\le\frac{C}{(1+\alpha^{p/2})(1+\beta^{p/2})}. $$ or $$ (1+\alpha^{p/2})(1+\beta^{p/2})\le C(1+\gamma^p). $$ This inequality is false. Let $x=(0,\dots,0,1)$, $y=(R,0,\dots,0)$ and $z=(-R,0,\dots,0)$. Then $\alpha=\beta=\sqrt{1+R^2}$ and $\gamma=1$. The left hand side goes t0 $\infty$ as $R\to\infty$, while the right hans side remans bounded.


Note

$\alpha$, $\beta$ and $\gamma$ have a geometrical interpretation. $\alpha$ and $\beta$ are the sides of a triangle, and $\gamma$ is the length of the median from the vertex fo the sides $\alpha$ and $\beta$ to the oposite side.

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Dear Professor Julián Aguirre, Thank you very much for your solutions. Even we restrict x,y,z to be real numbers, by letting x=1 and y=-z=R, the second inequality will fail as R tends to infinite. Thanks a lot. –  Anand Feb 24 '12 at 14:56
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