Inequality involvig complex numbers and their modulus If $a,b\in \mathbb{C}$ and $r&gt;1$, then $\left| 
 a|a|^{r-1}-b|b|^{r-1} \right|\leq{r\max\{|a|^{r-1},|b|^{r- 1}}\}|a-b|$.

If $a,b&gt;0$ this is a consequence of mean value theorem, but I don't know how to prove it in the general case. Is there a different version of this theorem for this case?

Thanks.
I hope this works.

Proposition:  if $ r \geqslant 1 $ and $ a,b \in \mathbb C $ then
    $$
  \left | a |a|^{r-1} - b |b|^{r-1} \right | \leqslant r \cdot \max\{ |a|^{r-1}, |b|^{r-1} \} \cdot |a - b|
 $$

Proof.  If either $a, b = 0 $ the inequality is trivially satisfied for all $ r \geqslant 1 $. We can assume without loss therefore that neither is zero and by symmetry we can arrange $ |a| \geqslant |b| &gt; 0 $.  

Write $ a = \alpha_1 e^{i r \theta_1 } $ and $ b = \alpha_2 e^{i r \theta_2} $.  Then the above ordering implies $\alpha_1 \geqslant \alpha_2 &gt; 0$.     

Write $\phi = \theta_2 - \theta_1 $ and define $m (r) $ by
    $$
  \begin{align}
   m &amp;= \left |a |a|^{r-1} - b |b|^{r-1} \right | \\
   &amp;= \left | \alpha_1^r e^{i\theta_1} - \alpha_2^r e^{i \theta_2} \right |
  \end{align}
 $$
whence,
    $$
  \begin{align}
   m^2 = \left | \alpha_1^r - \alpha_2^r e^{i\phi} \right |^2.
  \end{align}
 $$
Using real and complex parts, we find
    $$
  \begin{align}
   m^2 &amp;= (\alpha_1^r + \alpha_2^r cos \phi)^2 + (\alpha_2^r \sin\phi )^2 \\
    &amp;= \alpha_1^{2r} + \alpha_2^{2r} + 2\alpha_2^r\alpha_1^r \cos\phi
  \end{align}
 $$
In like fashion,
    $$
  \begin{align}
   |a - b|^2 &amp;= \left | \alpha_1 -\alpha_2 e^{i\phi} \right |^2 \\
    &amp;= \alpha_1^2 + \alpha_2^2 + 2 \alpha_1\alpha_2  \cos \phi 
  \end{align}
 $$
Now consider the function $ f $ of $ r $ for $ r \geqslant 1$ given by
    $$
  \begin{align}
   f(r) &amp;= r^2 \alpha_1^{2r-2} | a - b |^2 - m(r)^2 \\
    &amp;= r^2( \alpha_1^{2r} + \alpha_1^{2r-2}\alpha_2^2 + 2 \alpha_1^{2r-1}\alpha_2 \cos\phi ) 
     - m^2 .
  \end{align}
 $$
If we can show $f(r)$ is non-negative then $m^2 \leqslant r^2 \alpha_1^{2r-2} | a-b|^2 $ and the required results is proved.  Set $ \lambda = \alpha_2/\alpha_1 $ and note that $ 0 &lt; \lambda \leqslant 1 $.  Then,
    $$
  \begin{align}
  f(r) / \alpha_1^{2r} &amp;= r^2( 1 + \lambda^2 + 2\lambda \cos\phi) 
    - (1 + \lambda^{2r} + 2 \lambda^r \cos \phi ) \\
  &amp;= r^2(1+\lambda^2) - (1+ \lambda^{2r} ) + 2\lambda(r^2-\lambda^{r-1}) \cos\phi .
  \end{align}
 $$
Because $ r \geqslant 1 $ and $ \lambda \leqslant 1 $, the coefficient of $\cos\phi$ is non-negative, and if we replace $\cos \phi $ with its lowest value $-1$, the right hand side is not increased, and
    $$
  \begin{align}
   f(r)/\alpha_1^{2r} 
    &amp;\geqslant 
     r^2(1 + \lambda^2 -2\lambda) - (1+\lambda^{2r} - 2 \lambda^r) \\
    &amp;= r^2(1-\lambda)^2-(1-\lambda^r)^2 \tag 1.
  \end{align}
 $$
To conclude apply the mean value theorem to the difference $1-\lambda^r$ considered as a function of $\lambda$ on the interval $ [0,1] $.  Then for some $\xi, \lambda &lt; \xi &lt; 1 $
    $$
  1-\lambda^r = (1-\lambda)r\xi^{r-1} \leqslant r\cdot (1-\lambda)
 $$
Using this in (1) gives,
    $$
  f(r)/\alpha_1^{2r} \geqslant r^2(1-\lambda)^2 - r^2(1-\lambda)^2 = 0
 $$
Therefore $ f(r) $ is non-negative and the inequality is proved.
I hope this works.

Proposition:  if $ r \geqslant 1 $ and $ a,b \in \mathbb C $ then
    $$
  \left | a |a|^{r-1} - b |b|^{r-1} \right | \leqslant r \cdot \max\{ |a|^{r-1}, |b|^{r-1} \} \cdot |a - b|
 $$

Proof.  If either $a, b = 0 $ the inequality is trivially satisfied for all $ r \geqslant 1 $. We can assume without loss therefore that neither is zero and by symmetry we can arrange $ |a| \geqslant |b| &gt; 0 $.  

Write $ a = \alpha_1 e^{i r \theta_1 } $ and $ b = \alpha_2 e^{i r \theta_2} $.  Then the above ordering implies $\alpha_1 \geqslant \alpha_2 &gt; 0$.     

Write $\phi = \theta_2 - \theta_1 $ and define $m (r) $ by
    $$
  \begin{align}
   m &amp;= \left |a |a|^{r-1} - b |b|^{r-1} \right | \\
   &amp;= \left | \alpha_1^r e^{i\theta_1} - \alpha_2^r e^{i \theta_2} \right |
  \end{align}
 $$
whence,
    $$
  \begin{align}
   m^2 = \left | \alpha_1^r - \alpha_2^r e^{i\phi} \right |^2.
  \end{align}
 $$
Using real and complex parts, we find
    $$
  \begin{align}
   m^2 &amp;= (\alpha_1^r + \alpha_2^r cos \phi)^2 + (\alpha_2^r \sin\phi )^2 \\
    &amp;= \alpha_1^{2r} + \alpha_2^{2r} + 2\alpha_2^r\alpha_1^r \cos\phi
  \end{align}
 $$
In like fashion,
    $$
  \begin{align}
   |a - b|^2 &amp;= \left | \alpha_1 -\alpha_2 e^{i\phi} \right |^2 \\
    &amp;= \alpha_1^2 + \alpha_2^2 + 2 \alpha_1\alpha_2  \cos \phi 
  \end{align}
 $$
Now consider the function $ f $ of $ r $ for $ r \geqslant 1$ given by
    $$
  \begin{align}
   f(r) &amp;= r^2 \alpha_1^{2r-2} | a - b |^2 - m(r)^2 \\
    &amp;= r^2( \alpha_1^{2r} + \alpha_1^{2r-2}\alpha_2^2 + 2 \alpha_1^{2r-1}\alpha_2 \cos\phi ) 
     - m^2 .
  \end{align}
 $$
If we can show $f(r)$ is non-negative then $m^2 \leqslant r^2 \alpha_1^{2r-2} | a-b|^2 $ and the required results is proved.  Set $ \lambda = \alpha_2/\alpha_1 $ and note that $ 0 &lt; \lambda \leqslant 1 $.  Then,
    $$
  \begin{align}
  f(r) / \alpha_1^{2r} &amp;= r^2( 1 + \lambda^2 + 2\lambda \cos\phi) 
    - (1 + \lambda^{2r} + 2 \lambda^r \cos \phi ) \\
  &amp;= r^2(1+\lambda^2) - (1+ \lambda^{2r} ) + 2\lambda(r^2-\lambda^{r-1}) \cos\phi .
  \end{align}
 $$
Because $ r \geqslant 1 $ and $ \lambda \leqslant 1 $, the coefficient of $\cos\phi$ is non-negative, and if we replace $\cos \phi $ with its lowest value $-1$, the right hand side is not increased, and
    $$
  \begin{align}
   f(r)/\alpha_1^{2r} 
    &amp;\geqslant 
     r^2(1 + \lambda^2 -2\lambda) - (1+\lambda^{2r} - 2 \lambda^r) \\
    &amp;= r^2(1-\lambda)^2-(1-\lambda^r)^2 \tag 1.
  \end{align}
 $$
To conclude apply the mean value theorem to the difference $1-\lambda^r$ considered as a function of $\lambda$ on the interval $ [0,1] $.  Then for some $\xi, \lambda &lt; \xi &lt; 1 $
    $$
  1-\lambda^r = (1-\lambda)r\xi^{r-1} \leqslant r\cdot (1-\lambda)
 $$
Using this in (1) gives,
    $$
  f(r)/\alpha_1^{2r} \geqslant r^2(1-\lambda)^2 - r^2(1-\lambda)^2 = 0
 $$
Therefore $ f(r) $ is non-negative and the inequality is proved.