An inequality involving $|a|^p$: how can I prove this?

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 $$C_L = \max(C_{L0},C_{L1},C_{L\infty},C_{L1-},C_{L1+}) = C_{L1+} = \color{brown}{\dfrac{1+\varepsilon}{\left(\sqrt[p-1]{1+\varepsilon}-1\right)^{p-1}}}.$$