Is it true that $||a|^p - |b|^p| \le C_p|a - b|^p$ for $1 < p < \infty$? If $a, b \in \mathbb{C}$, then we have the standard triangle inequality for the difference:
$$||a| - |b|| \le |a - b|.$$
I am wondering if this inequality generalizes to exponents greater that one.

My question is, for $1 < p < \infty$, does there exists a constant $C_p$ such that $||a|^p - |b|^p| \le C_p|a - b|^p$ for all $a, b \in \mathbb{C}$?

I am aware of the "standard" triangle inequality for $p > 1$:
$$|a +b|^p \le 2^{p}(|a|^p + |b|^p),$$
and that $2^p$ may not be sharpest constant possible. If I try to use this estimate to prove, say, that $|a|^p - |b|^p \le C_p |a - b|^p$, I get stuck with an extra term that I'm not sure what to do with.
$$|a|^p - |b|^p \le 2^p|a -b|^p + (2^p -1)|b|^p.$$
A resolution on this matter is greatly appreciated.
 A: Hint: Is $(x+1)^2 -x^2$ bounded on $(0,\infty)?$
A: This is false for $p>1$. In fact, by the mean value theorem, there is for $n\in \Bbb {N} $ some $\xi_n \in (n,n+1) $ with
$$
(n+1)^p - n^p = p \cdot \xi_n^{p-1}\to \infty 
$$
as $n\to \infty $. But if your claim was true, the left-hand side would be bounded by $C_p $.

EDIT: For $p < 1$, the inequality is true.
To see this, first note that $f(x) := (1 + x)^p \leq 1 + x^p =: g(x)$ for all $x \in [0,\infty)$.
Indeed, $f,g : [0,\infty) \to [1,\infty)$ are continuous with
$f(0) = 1 = g(0)$ and that $f'(x) = p \cdot (1 + x)^{p-1} \leq p \cdot x^{p-1}$ since $p < 1$.
Next, note that if $a,b > 0$ and $x := a/b$, then multiplying the inequality $f(x) \leq g(x)$
by $b^p$ leads to $(b + a)^p \leq b^p + a^p$ for all $a,b > 0$.
For $b = 0$ or $a = 0$, this trivially continues to hold.
Finally, for $a,b \in \Bbb{C}$, we thus see
$$
  |a|^p
  = |(a - b) + b|^p
  \leq (|a-b| + |b|)^p
  \leq |a-b|^p + |b|^p
$$
and hence $|a|^p - |b|^p \leq |a-b|^p$.
By symmetry (swap $a,b$), this implies $\big| |a|^p - |b|^p \big| \leq |a-b|^p$ as desired.
