An inequality involving $|a|^p$: how can I prove this? How can I prove the following lemma?

Let $1<p<\infty$ and let $\epsilon >0$. Then there exists a constant $C \geq 0$ (may depend on $p$ and $\epsilon$) such that for all $a, b \in \mathbb{C}$, $||a+b|^p - |b|^p| \leq \epsilon |b|^p +C|a|^p $.

I am suffering since I cannot use any homogeneity argument because of the $|a+b|^p$ term. I tried to use the fact that $|t|^p$ is a convex function for each fixed $p>1$, but it does not seem to work, as far as I tried.
Thanks!
 A: Let 
$$x =|a+b|,\quad y = |b|,$$
then the issue inequality takes the form of
$$|x^p-y^p| \le \varepsilon y^p + C|x+e^{i\varphi} y|^p,\tag1$$
where the phase $\varphi$ depends of the phases of $(a+b)$ and $b.$
Since for the arbitrary $u,v\in\mathbb C$
\begin{cases}
||u|-|v|| \le |u+v|\\
||u|-|v|| \le |u-v|,
\end{cases}
then the worst case of $(1)$ is the case
$$|x^p-y^p| \le \varepsilon y^p + C|x - y|^p.\tag2$$
Inequality $(2)$ is homogenuis by $x$ and $y,$ so the least value of $C$ should provide the inequality
$$|z^p-1| \le \varepsilon + C|z - 1|^p,\tag3$$
where $z\in[0,\infty).$
If $\color{brown}{z=0},$ then $C_{L0} =1-\varepsilon.$
If $\color{brown}{z=1},$ then $C_{L1} =0.$
If $\color{brown}{z\to \infty},$ then $C_{L\infty} = 1.$
Denote 
$$f(C,z) = |z^p-1| - \varepsilon - C|z - 1|^p.\tag4$$
The least value of $C$ can be defined from the condition
$$\max\limits_{z\in[0,\infty)} f(C,z) = 0.$$
The inner maxima can be achived only if $f'_z(C,z)=0.$
If $\color{brown}{z\in(0,1)}$ then
$$f(C,z) = 1-z^p-\varepsilon - C(1-z)^p,$$
\begin{cases}
-pz^{p-1}+Cp(1-z)^{p-1} = 0\\
1-z^p-\varepsilon - C(1-z)^p = 0,
\end{cases}
\begin{cases}
\dfrac z{1-z}=\sqrt[p-1]C,\quad z = \dfrac{\sqrt[p-1]C}{\sqrt[p-1]C+1}\\
1 - \varepsilon = \left(\dfrac{\sqrt[p-1]C}{\sqrt[p-1]C+1}\right)^p
+ C\left(\dfrac{1}{\sqrt[p-1]C+1}\right)^p,
\end{cases}
$$1 - \varepsilon = \dfrac{C\left(\sqrt[p-1]C+1\right)}{\left(\sqrt[p-1]C+1\right)^p}
 = \dfrac{C}{\left(\sqrt[p-1]C+1\right)^{p-1}} = z^{p-1},\quad z=\sqrt[p-1]{1-\varepsilon},$$
$$C_{L1-}= \left(\dfrac z{1-z}\right)^{p-1} = \dfrac{1-\varepsilon}{\left(1-\sqrt[p-1]{1-\varepsilon}\right)^{p-1}}.$$
If $\color{brown}{z\in(1,\infty)}$ then
$$f(C,z) = z^p-1-\varepsilon - C(z-1)^p,$$
\begin{cases}
pz^{p-1}-Cp(z-1)^{p-1} = 0\\
z^p-1-\varepsilon - C(z-1)^p = 0,
\end{cases}
\begin{cases}
\dfrac z{z-1}=\sqrt[p-1]C,\quad z = \dfrac{\sqrt[p-1]C}{\sqrt[p-1]C-1}\\
1 + \varepsilon = \left(\dfrac{\sqrt[p-1]C}{\sqrt[p-1]C-1}\right)^p
- C\left(\dfrac{1}{\sqrt[p-1]C-1}\right)^p,
\end{cases}
$$1 + \varepsilon = \dfrac{C\left(\sqrt[p-1]C-1\right)}{\left(\sqrt[p-1]C-1\right)^p}
 = \dfrac{C}{\left(\sqrt[p-1]C-1\right)^{p-1}} = z^{p-1},\quad z=\sqrt[p-1]{1+\varepsilon},$$
$$C_{L1+}= \left(\dfrac z{z-1}\right)^{p-1} = \dfrac{1+\varepsilon}{\left(\sqrt[p-1]{1+\varepsilon}-1\right)^{p-1}}.$$
Since
$$\dfrac{C_{L1-}}{C_{L1+}}
 = \dfrac{1-\varepsilon}{1+\varepsilon}\left(\dfrac{\sqrt[p-1]{1+\varepsilon}-1}{1-\sqrt[p-1]{1-\varepsilon}}\right)^{p-1}
 = \left(\dfrac{\sqrt[p-1]{1-\varepsilon}}{\sqrt[p-1]{1+\varepsilon}}\dfrac{\sqrt[p-1]{1+\varepsilon}-1}{1-\sqrt[p-1]{1-\varepsilon}}\right)^{p-1}$$
$$ = \left(\dfrac{\sqrt[p-1]{1-\varepsilon^2}-\sqrt[p-1]{1-\varepsilon}}{\sqrt[p-1]{1+\varepsilon}-\sqrt[p-1]{1-\varepsilon^2}}\right)^{p-1}
 = \left(1+\dfrac{2\ \sqrt[p-1]{1-\varepsilon^2}-\sqrt[p-1]{1-\varepsilon}-\sqrt[p-1]{1+\varepsilon}}{\sqrt[p-1]{1+\varepsilon}-\sqrt[p-1]{1-\varepsilon^2}}\right)^{p-1} \le1,$$
then the least value of the constant $C$ is
$$C_L = \max(C_{L0},C_{L1},C_{L\infty},C_{L1-},C_{L1+}) = C_{L1+}
 = \color{brown}{\dfrac{1+\varepsilon}{\left(\sqrt[p-1]{1+\varepsilon}-1\right)^{p-1}}}.$$
A: Let $\epsilon>0$ be fixed. The inequality is trivial for $b=0$. Else, setting $t=a/b\in\mathbb C$, we are asked to prove $$||1+t|^p-1|\le \epsilon +C_{\epsilon,p} |t|^p. \label{*}\tag{*}$$
On the set $|t|\ge c>0$, $c>0$ to be determined, note that by convexity of $|t|^p$,
$$\frac{||1+t|^p-1| }{ |t|^p}\le \frac{2^{p-1}+2^{p-1}|t|^p+1}{|t|^p} = \frac{2^{p-1}+1}{|t|^p} + 2^{p-1} \le \frac{2^{p-1}+1}{c^p} + 2^{p-1} =: M_{c,p}. $$ 
Thus, we have for any $c$
$$ ||1+t|^p-1| \le M_{c,p} |t|^p \le  \epsilon  + M_{c,p} |t|^p.$$
On the set $|t|<c$, we have by the Mean Value Inequality and $\mathbb R^2$-differentiability of $|s|^p$ at $s=1$, (writing $[1,1+t]$ for the line segment in $\mathbb R^2\cong \mathbb C$)
$$ ||1+t|^p - 1| \le \sup_{s\in[1,1+t]} p|s|^{p-1} |t| \le p(1+|t|)^{p-1}|t| \le p(1+|c|)^{p-1}|c|. $$
Now choose $c=c(\epsilon,p)$ so small that
$$ p(1+|c|)^{p-1}|c|< \epsilon,$$
giving
$$ ||1+t|^p - 1| \le \epsilon \le \epsilon  + M_{c(\epsilon,p),p} |t|^p$$
We have now proven (\ref{*}), with $C_{\epsilon,p} = M_{c(\epsilon,p),p}$.
P.S. $C_{\epsilon,p}$ necessarily explodes as $\epsilon\to 0$, see here and here.
