An inequality from PDEs: $|v|^\alpha v -|w|^\alpha w \leq C(|v|^\alpha + |w|^\alpha)|v-w|$ In using Stritcharz estimates to prove well-posedness of the nonlinear Schrödinger equation
$$i\partial_tu = \Delta u - \lambda|u|^\alpha u$$
where $\alpha>0$, one requires the following inequality for complex numbers $v,w$:
$$|v|^\alpha v -|w|^\alpha w \leq C(|v|^\alpha + |w|^\alpha)|v-w|.$$
See, for instance, p. 84 of Semilinear Schrödinger Equations by Cazenave.
When $x$ and $y$ are real and positive, I can see the result as a consequence of the mean value theorem. Applied to the map $x \mapsto x^{\alpha+1}$, for each $0<x<y$ there exists $\xi\in[x,y]$ such that
$$\frac{y^{\alpha+1}-x^{\alpha+1}}{y-x} = (\alpha+1)\xi^\alpha \leq (\alpha+1)(x^\alpha + y^\alpha).$$
The inequality follows as $x \mapsto x^\alpha$ increasing implies $\xi^\alpha \leq \max\{x^\alpha,y^\alpha\}$.
I've thought about this on-and-off for the last month thinking I'd figure out the more general result, but that eventuality never occurred. Am I missing something simple? Thanks!

Forget complex structure and work with $\mathbb R^2$, or better yet $\mathbb R^n$. For any differentiable map $F:\mathbb R^n\to\mathbb R^n$ we have the Mean Value Inequality $$|F(u)-F(v)|\le |u-v|\max_{[u,v]}\|DF\| \tag1$$ where $|DF|$ is the operator norm of the derivative matrix, and $[u,v]$  is the line segment with endpoints $u,v$. If you want to prove (1) yourself, apply the 1-variable mean value theorem to the function $F(tu+(1-t)v)$.
Here $F(u)=|u|^\alpha u$. It remains to prove the estimate $\|DF(u)\|\le C|u|^\alpha$. The laziest thing to do is to let $C=\max_{|u|=1}\|DF(u)\|$, which is finite by continuity/compactness. Then appeal to homogeneity: $$F(\lambda u)=\lambda^{1+\alpha}F(u)\tag2$$ for all $\lambda>0$ and all $u\in\mathbb R^n$, Differentiate both sides of (2) using  the Chain Rule: $$\lambda DF(\lambda u)=\lambda^{1+\alpha}DF(u)$$ Hence $DF(\lambda u)=\lambda^{\alpha}DF(u)$, which implies $|DF(u)|\le C|u|^\alpha $ by the definition of $C$.
