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 $z_i \in \mathbb C$.
 A: Here's a more general result: For $1\le p < \infty$ and $z,w\in \mathbb {C},$
$$|\,|z|^pz - |w|^pw|\ | \le p|z-w|(|z|^p+|w|^p).$$
Proof: WLOG $|z|< |w|,$ the case $|z|=|w|$ being easy. We have 
$$\tag 1  |z|^pz - |w|^pw =  |z|^p(z - w) + (|z|^p - |w|^p)w.$$
Claim: $|\ |z|^p - |w|^p \ | \le  p|w|^{p-1}|z-w|.$ Assuming the claim, take absolute values in $(1)$ and use the triangle inequality to see
$$|\,|z|^pz - |w|^pw\,| \le |z|^p|z - w|+p|w|^{p}|z-w|\le p|z-w|(|z|^p + |w|^p),$$
which is the desired conclusion. To prove the claim, use the MVT to see
$$\tag 2 |z|^p - |w|^p = pc^{p-1}(|z|-|w|)$$
where $|z|<c<|w|.$ Because $p-1\ge 0,$ and $|z|<|w|,c^{p-1} < |w|^{p-1}.$ Thus $(2)$ implies
$$ ||z|^p - |w|^p | \le p|w|^{p-1}||z|-|w|| \le p|w|^{p-1}|z-w|.$$
A: 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 $2C \geq 3$.
