Triangle inequality: $\vert\vert a+b\vert^q-\vert a\vert^q\vert\leq \varepsilon\vert a\vert^q+C(\varepsilon)\vert b\vert^q$ I am having difficulty in proving the following inequality for $1\leq q<\infty$:
\begin{equation}
\Big|\vert a+b\vert^q-\vert a\vert^q\Big|\leq \varepsilon\vert a\vert^q+C(\varepsilon)\vert b\vert^q\quad (a, b\in\mathbb{R}, \varepsilon>0)
\end{equation}
where $C(\varepsilon)$ depends only on $\varepsilon$ and $q$.
I thought it might have something to do with:
\begin{equation}
\vert a+b\vert^q\leq 2^{q-1}(\vert a\vert^q+\vert b\vert^q)
\end{equation}
(where I use an appropriate substitution) but I couldn't get anywhere.
Then I tried cases because we know that:
\begin{equation}
\Big|\vert a+b\vert^q-\vert a\vert^q\Big|\leq (2^{q-1}+1)\vert a\vert^q+2^{q-1}\vert b\vert^q\quad (a, b\in\mathbb{R}, \varepsilon>0)
\end{equation}
so if $\varepsilon\geq 2^{q-1}+1$ then we can set $C(\varepsilon)=\varepsilon-1$. The issue is if $\varepsilon<2^{q-1}+1$ then ofcourse it doesn't work.
 A: Let's take cases. Let $a, b\in \mathbb{R}$.
Note that the function $t\mapsto t^q$ is convex $(t\geq 0),\ q\geq 1$.
Case 1; q>1.
Given $\varepsilon>0$ define
\begin{equation}
\eta\equiv(1+\varepsilon)^{\frac{1}{q-1}}-1.
\end{equation}Then
\begin{equation}
\lambda\equiv\frac{1}{1+\eta}\in (0, 1)
\end{equation}
and by the convexity of $t^q$ we obtain
\begin{align}
|a+b|^q-|a|^q\leq(|a|+|b|)^q-|a|^q&=\left(\lambda\frac{|a|}{\lambda}+(1-\lambda)\frac{|b|}{1-\lambda}\right)^q-|a|^q\\
&\leq\lambda^{1-q}|a|^q+(1-\lambda)^{1-q}|b|^q-|a|^q\\
&=[\lambda^{1-q}-1]|a|^q+(1-\lambda)^{1-q}|b|^q.
\end{align}
Therefore,
\begin{align}
\big||a+b|^q-|a|^q\big|&\leq |\lambda^{1-q}-1||a|^q+(1-\lambda)^{1-q}|b|^q\\
&=|(\eta+1)^{q-1}-1||a|^q+\left(1+\frac{1}{\eta}\right)^{q-1}|b|^q\\
&=\varepsilon|a|^q+\left(1+\frac{1}{(1+\varepsilon)^{\frac{1}{q-1}}-1}\right)|b|^q\\
&\equiv \varepsilon|a|^q+C(\varepsilon)|b|^q.
\end{align}
Case 2; q=1. Given $\varepsilon>0$ set $C(\varepsilon)\equiv 1$, then
\begin{equation}
|a+b|-|a|\leq |a|+|b|-|a|=|b|\leq \varepsilon|a|+|b|
\end{equation}and hence
\begin{equation}
\big||a+b|^q-|a|^q\big|\leq \varepsilon|a|+C(\varepsilon)|b|
\end{equation}
