Peetre’s inequality

Theorem  [Peetre’s inequality] [1, 2]If $t$ is a real number and $x,y$ are vectors in${ℝ}^n$, then

$${\left(\frac{1+{|x|}^2}{1+{|y|}^2}\right)}^t\le 2^{|t|}{(1+{|x-y|}^2)}^{|t|}.$$

Proof. (Following [1].)Suppose $b$ and $c$ are vectors in ${ℝ}^n$. Then, from${(|b|-|c|)}^2\ge 0$, we obtain

$$2|b|\cdot |c|\le {|b|}^2+{|c|}^2.$$

Using this inequality and the Cauchy-Schwarz inequality, we obtain

$1+{|b-c|}^2$$=$$1+{|b|}^2-2b\cdot c+{|c|}^2$

$\le$$1+{|b|}^2+2|b||c|+{|c|}^2$

$\le$$1+2{|b|}^2+2{|c|}^2$

$\le$$2\left(1+{|b|}^2+{|c|}^2+{|b|}^2{|c|}^2\right)$

$=$$2(1+{|b|}^2)(1+{|c|}^2)$

Let us define $a=b-c$. Then for any vectors $a$ and $b$, we have

${\displaystyle \frac{1+{|a|}^2}{1+{|b|}^2}}\le 2(1+{|a-b|}^2).$(1)

Let us now return to the given inequality.If $t=0$, the claim is trivially true for all $x,y$ in ${ℝ}^n$.If $t>0$, then raising both sides in inequality 1 tothe power of $t$, using $t=|t|$, and setting $a=x$, $b=y$ yields the result.On the other hand, if , then raising both sides in inequality1 to the power to $-t$, using $-t=|t|$, and setting$a=y$, $b=x$ yields the result.$\mathrm{\square }$