$|x+y|^p \leq |x|^p + |y|^p$ for $0 < p < 1$. How can I prove the inequality $|x+y|^p \leq |x|^p+ |y|^p$ for $0 < p < 1$ and $x,y \in \mathbb{C}$. 
I have an idea when $x,y \in \mathbb{R}$: then I can prove the equivalent
$$(1+s)^p \leq 1+ s^p$$
where $s > 0$ but for the complex case I don't really have ideas.
Thanks for the help/reference!
 A: For $x=y=0$ the inequality (or should I say equality) holds. Now we are going to prove it also holds for at least one non - zero $x$ or $y$.
$a^{p} \geq a$ for $0<p<1$ and $0 \leq a \leq 1$ where equality happens if $a$ is either $0$ or $1$.
$\left( \frac{|x|}{|x|+|y|} \right)^{p} \geq \frac{|x|}{|x|+|y|}$
$\left( \frac{|y|}{|x|+|y|} \right)^{p} \geq \frac{|y|}{|x|+|y|}$
Adding the two inequalities, we obtain
$\left( \frac{|x|}{|x|+|y|} \right)^{p} + \left( \frac{|y|}{|x|+|y|} \right)^{p} \geq 1$
$|x|^{p} + |y|^{p} \geq (|x|+|y|)^{p}$
Combine with triangle inequality $|x|+|y| \geq |x+y|$ to obtain 
$|x|^{p} + |y|^{p} \geq |x+y|^{p}$, equality happens if any of $x,y$ is zero.
A: $x\mapsto x^p$ is increasing, so $|x+y|^p\leq (|x|+|y|)^p$ by the triangle inequality, so all you have to prove is that $(a+b)^p \leq a^p+b^p$ for all $a,b\geq 0$.
To see this, fix $b$ and define $g(x)=x^p+b^p-(x+b)^p$ and note that $g'(x)=px^{p-1}-p(x+b)^{p-1}\geq 0,$ so $g$ is increasing. We also see that $g(0)=0$ and therefore $g(x)\geq 0$. This establishes the desired result.
