Peetre's Inequality - not strict? 
(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*}

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)

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*}
