Looking for hints of this inequality 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
 A: 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.
