Is it true that $|a^{\alpha} - b^{\alpha}| \leq |a-b|^{\alpha}$? I am currently reading some papers that seem to use the fact that $$|a^{\alpha} - b^{\alpha}| \leq |a-b|^{\alpha},$$ for $-1< \alpha < 0$ and $a,b$ in the upper half plane of $\mathbb{C}$. Is this true or not? I want to use differentiation to prove this, but I don't know how to deal with this situation in the complex case. Any comments?
 A: For the new version of the problem, with $-1 < \alpha < 0$, consider $a = 1$, $b = -1$.  We have 
$$ \left| \dfrac{1^{-1} - (-1)^{-1}}{(1 - (-1))^{-1}}\right| = 4$$
so for $\alpha \to -1+$ with $b \to -1$ and $a \to +1$ in the upper half plane, 
$$ \dfrac{|a^\alpha - b^\alpha|}{|a - b|^\alpha} = 
\left| \dfrac{a^\alpha - b^\alpha}{(a-b)^\alpha}\right| \to 4$$
For example, with $b = -1 + i/10$, $a = 1 + i/10$, $\alpha = -9/10$, 
$$\dfrac{|a^\alpha - b^\alpha|}{|a - b|^\alpha} \approx 3.6 > 1$$
A: Edit: OP originally asked if the inequality is true for $\alpha\in(0,1)$. It is true if $\alpha\in(0,1)$. The inequality is equivalent to
$$ |x^\alpha-1|\le|x-1|^\alpha. $$
W.L.OG, let $x\ge 1$. Define
$$ f(x)=(x^\alpha-1)-(x-1)^\alpha, x\in[1,\infty). $$
Then
$$ f'(x)=\alpha [x^{\alpha-1}-(x-1)^{\alpha-1}]=\alpha\left[\frac{1}{x^{1-\alpha}}-\frac{1}{(x-1)^{1-\alpha}}\right]. $$
Noting that $x>x-1>0$ implies $\frac{1}{x}<\frac{1}{x-1}$, we have $f'(x)<0$ since $1-\alpha>0$. This implies that $f(x)$ is strictly decreasing. So for $x\ge1$, we have $f(x)\le f(0)=0$ or
$$ |x^\alpha-1|\le|x-1|^\alpha. $$
A: A good answer is given for $a,b$ real. I don't think it holds for all complex numbers, but it might hold in some region of the complex plane. Please let me know if the following has any errors.
The inequality is equivalent to
$$|c^\alpha - 1| \leq |c-1|^\alpha.$$
Taking $c=|c|e^{i\theta}$ we can write this as
$$||c|^\alpha - e^{-i a \theta}| \leq ||c|-e^{-i \theta}|^\alpha$$
Choose $\alpha \theta = \pi$, $\alpha = 1/2$:
$$||c|^\alpha + 1| \leq ||c|-e^{-i \pi/a}|^\alpha$$
$$|c^{1/2}+1| \leq |c-1|^{1/2}$$
$$c+2c^{1/2}+1 \leq c-1$$
which can't be true for all $c$.
