Peetre's Inequality - not strict?

Question

(Peetre's inequality) Let $x,y \in \Bbb R^n$ and $s \in \Bbb R$. Then $$ \frac{(1+|x|^2)^s}{(1+|y|^2)^s} \le 2^{|s|} (1+|x-y|^2)^{|s|}.$$

Proof: By switching roles of $x,y$ we may suppose $s \ge 0$, and taking $s$th root may assume $s=1$. Then the argument i found online:

\begin{align*} (1+|x|^2) &= 1 + |x-y|^2 + |y|^2 +2(x-y)y \\ & \le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \\ & \le 2(1+|y|^2 + |x-y|^2 +|y|^2|x-y|^2 ) \\ & = 2(1+|y|^2)(1+|x-y|^2). \end{align*}

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What I don't understand is that on the third inequality, isn't this clearly a strict inequality when $s\not= 0$? (at least by 1)

Answer 1

Yes, you are correct. We have that for $x,y\in \mathbb{R}^n$, \begin{align*} (1+|x|^2) &=1+|(x-y)+y|^2\\ &= 1 + |x-y|^2 + |y|^2 +2\langle (x-y), y \rangle \\ &\leq 1 + |x-y|^2 + |y|^2 +2|x-y| |y| \\ & \le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \\ & = 1+ 2|y|^2 + 2|x-y|^2 \\ & < 2+ 2|y|^2 +2|x-y|^2 +2|y|^2|x-y|^2 \\ & = 2(1+|y|^2)(1+|x-y|^2). \end{align*}