Proof of the inequality $||a+b|^p - |a|^p| \leq \epsilon |a|^p + C_{\epsilon} |b|^p$

Define $k=\dfrac{b}{a}$. This theorem is equivalent to show that $$\left||1+k|^p-1\right|\le\epsilon+C_\epsilon|k|^p$$First let $|1+k|\ge 1$, then we must have$$|1+k|^p-1\le\epsilon+C_\epsilon|k|^p$$or equivalently$$\dfrac{|1+k|^p-1-\epsilon}{|k|^p}\le C_\epsilon$$if $|1+k|^p\le1+\epsilon$ the theorem holds with any $C_\epsilon>0$ then assume that $|1+k|^p>1+\epsilon$. By triangle inequality we have $$(1+|k|)^p\ge|1+k|^p>1+\epsilon$$which yields to $$|k|>(1+\epsilon)^{\frac{1}{p}}-1$$or$$|\dfrac{1}{k}|<\dfrac{1}{(1+\epsilon)^{\frac{1}{p}}-1}$$therefore $$\dfrac{|1+k|^p-1-\epsilon}{|k|^p}<(1+|\dfrac{1}{k}|)^p<\dfrac{1+\epsilon}{((1+\epsilon)^{\frac{1}{p}}-1)^p}$$then by choosing $C_\epsilon=\dfrac{1+\epsilon}{((1+\epsilon)^{\frac{1}{p}}-1)^p}$ we complete our proof in this case.

The other case is where $|1+k|<1$. Here we need to show that $$\dfrac{1-|1+k|^p-\epsilon}{|k|^p}\le C_\epsilon$$ for $|1+k|^p\ge 1-\epsilon$ this automatically holds and for $|1+k|^p<1-\epsilon$ we have $$|1+k|<(1-\epsilon)^{\dfrac{1}{p}}$$this is inside a circle of radius $(1-\epsilon)^{\dfrac{1}{p}}$ centered at $-1$ which geometrically means that $$|k|>1-(1-\epsilon)^{\dfrac{1}{p}}$$. Here taking $C_\epsilon=\dfrac{1}{\left(1-(1-\epsilon)^{\dfrac{1}{p}}\right)^p}$ fulfills our condition. Therefore the general $C_\epsilon$ would be $$\Large C_\epsilon=\max\left\{\dfrac{1}{\left(1-(1-\epsilon)^{\frac{1}{p}}\right)^p},\dfrac{1+\epsilon}{((1+\epsilon)^{\frac{1}{p}}-1)^p}\right\}$$