Complex inequality $||u|^{p-1}u – |v|^{p-1}v|\leq c_p |u-v|(|u|^{p-1}+|v|^{p-1})$

How does one show for complex numbers u and v, and for p>1 that

||u|^{p-1}u – |v|^{p-1}v|\leq c_p |u-v|(|u|^{p-1}+|v|^{p-1}),

where $c_p$ is some constant dependent on p. My intuition is to use some version of the mean value theorem with $F(u) = |u|^{p-1}u$, but I’m not sure how to make this work for complex-valued functions. Plus there seems to be an issue with the fact that $F$ may not smooth near the origin.

For context, this shows up in Terry Tao’s book Nonlinear Dispersive Equations: Local and Global Analysis on pg. 136, where it is stated without proof as an “elementary estimate”.

Solutions Collecting From Web of "Complex inequality $||u|^{p-1}u – |v|^{p-1}v|\leq c_p |u-v|(|u|^{p-1}+|v|^{p-1})$"

Suppose without loss of generality that $|u| \geq |v| > 0$. Then you can divide the equation through by $|v|^p$ and your task it to prove $||w|^{p-1}w – 1| \leq c_p|w – 1|(|w|^{p-1} + 1)$, where $w = u/v$. Note that
$$||w|^{p-1}w – 1| = ||w|^{p-1}w – |w|^{p-1} + |w|^{p-1} – 1| $$
$$\leq ||w|^{p-1}w – |w|^{p-1}| + ||w|^{p-1} – 1|$$
Note the first term is $|w|^{p-1}|w – 1|$ is automatically bounded by your right hand side. So you’re left trying to show that $||w|^{p-1} – 1|$ is bounded by your right hand side. For this it suffices to show that
$$||w|^{p-1} – 1| \leq c_p||w| – 1|| (|w|^{p-1} + 1)$$
Since $|w| \geq 1$ by the assumption that $|u| \geq |v|$, it suffices to show
that for all real $r \geq 1$ one has
$$r^{p-1} – 1 \leq c_p(r – 1)(r^{p-1} + 1)$$
Now use the mean value theorem as you originally wanted to.

Without loss, you can assume $|u|\le|v|$ and replace $u$ by $uv$ to reduce the problem to the situation $v=1$ and $|u|\le1$. Unless you need $c_p$ explicitly, then it’s clear, yes?