TITLE: Show $\frac{1}{C} \langle t + a_k \rangle \le \langle t+ b_k \rangle \leq C\langle t+ a_k\rangle$
QUESTION [0 upvotes]: Let $\langle * \rangle = (1+|*|^2)^{1/2}$ denote the Japanese packets. Let $a_k$ and $b_k$ are two real sequences. If there is $C>0$ such that
$$|a_k -b_k| < C$$
then there is $C>0$ such that $\forall k \in \mathbb N$, $\forall t \in \mathbb R$ we have
$$\frac{1}{C} \langle t + a_k \rangle \le \langle t+ b_k \rangle \leq  C\langle  t+ a_k\rangle$$
I spend a lot of time but could not get a reasonable proof. Any help is appreciated.

REPLY [1 votes]: Hints:

$\langle\cdot\rangle$ is Lipschitz continuous with constant $1$,
$\langle x \rangle \ge 1$ for all $x \in \mathbb R$.
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