I'm wondering if the following 'additive' Peetre's inequality has its own name?
for $x,y\in\mathbb R^d$, $p\ge 0$, and $c=\max(2^p,2^{2p-1})$, $$|1+|x|^2|^p \leq c \big((1+|x-y|^2)^p + (1+|y|^2)^p\big)$$
If not a name, then a reference where someone uses it will do...
I just recently found out that Peetre's inequality is instead usually the following 'multiplicative' inequality, valid for all $s\in\mathbb R$ and $x,y\in\mathbb R^d$:
\begin{equation} (1 + |x + y|^2)^s \leq 2^{|s|}(1 + |x|^2)^s(1 + |y|^2)^{|s|} \end{equation} There's also this related inequality Showing the inequality $|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)$ .
I'll end the question with a proof of the boxed inequality just in case someone is worried.
For $p>1/2$, the function $\phi(x)=|1+|x|^2|^p$ is convex. Thus $\phi(x/2) \le \frac{\phi(y)}2 + \frac{\phi(x-y)}2$, i.e. $$ |1+|x/2|^2|^p \leq \frac12 \big((1+|x-y|^2)^p + (1+|y|^2)^p\big)$$ Multiplying by $2^{2p}$ we obtain $$ |4+|x|^2|^p \leq 2^{2p-1} \big((1+|x-y|^2)^p + (1+|y|^2)^p\big),$$ which is clearly enough. For $p<1$, instead use the convexity of $1+|x|^2$, getting $$4 +|x|^2 \le 2 (1+|y|^2) + 2(1+|x-y|^2),$$ and then the subadditivity of $|\cdot|^p$ for $p<1$ finishes the proof.
