Ineqaulity: $\left| |z_1|^2 z_1 - |z_2|^2 z_2 \right| \le C(|z_1|^2 + |z_2|^2)|z_1 - z_2|$? How can I prove the following inequality:
$$
\left| |z_1|^2 z_1 - |z_2|^2 z_2 \right|
\le
C(|z_1|^2 + |z_2|^2)|z_1 - z_2|
$$
with some constant $C>0$?
Here $z_i \in \mathbb C$.

For shortness, let $x = z_1, y = z_2$.
$$
\big||x|^2 x - |y|^2 y\big| \leq 
\big||x|^2 x - |x|^2 y\big| +
\big||y|^2 x - |x|^2 y\big| +
\big||y|^2 x - |y|^2 y\big| =
|x|^2 |x-y| + |y|^2 |x-y| + \big||y|^2 x - |x|^2 y\big|
$$
Note that
$$
\big||y|^2 x - |x|^2 y\big| = 
\big|xy\bar{y} - xy\bar{x}\big| = |xy||\bar x-\bar y| = |x||y||x-y|.
$$
Since $|xy| \leq \frac{|x|^2 + |y|^2}{2}$ due to AM-GM inequality for $|x|^2$ and $|y|^2$,
$$
\big||x|^2 x - |y|^2 y\big| \leq \frac{3}{2}\left(|x|^2 + |y|^2\right) |x-y|.
$$
The bound is tight, for example consider $x = 1, y = 1 + \delta, \delta > 0$.
$$
\big||x|^2 x - |y|^2 y\big| = (1 + \delta)^3 - 1 = 3 \delta + O(\delta^2)\\
C\left(|x|^2 + |y|^2\right) |x-y| = C\left((1+\delta)^2 + 1\right) \delta = 2C\delta + O(\delta^2),
$$
thus $2C \geq 3$.
