Inequality involving sum and modulus of complex numbers Let $z_1,z_2,...,z_n$ be complex numbers, prove the inequality:
$$\frac{\vert\sum_{k=1}^n z_k\vert}{1+\vert\sum_{k=1}^n z_k\vert} \le \sum_{k=1}^n \frac{\vert z_k \vert}{1+\vert z_k \vert}$$
 A: Let $f(z)=\frac{\vert z \vert}{1+\vert z \vert}=1-\frac{1}{1+\vert z \vert}$, then rewriting your inequality we get : 
$$f\left(\sum_{k=1}^n z_k\right) \leq \sum_{k=1}^n f(z_k).$$
If you prove the statement for $n=2$, then every thing is proved by induction;
could you continue it yourself?
A: For the base case $n=2$, let $a= \mid z_1 \mid $ and $b=\mid z_2 \mid $
\begin{eqnarray*}
 &ab(a+b+2) &  \geq &  0 \\
 & ab(a+b)+b(2a+b) +b & \\ &+ab(a+b)+a(a+2b) +a &  \geq & ab(a+b)+a(a+b)+b(a+b) +a+b  \\
&a(1+b)(1+a+b)+b(1+a)(1+a+b) &  \geq &  (a+b)(1+a)(1+b) \\
& \frac{a}{1+a}+\frac{b}{1+b}   \geq  \frac{a+b}{1+a+b}
\end{eqnarray*}
The inductive step is established by ...
\begin{eqnarray*}
\sum_{k=1}^{n+1} \frac{\vert z_k \vert}{1+\vert z_k \vert} &=&\sum_{k=1}^{n} \frac{\vert z_k \vert}{1+\vert z_k \vert} +\frac{\vert z_{n+1} \vert}{1+\vert z_{n+1} \vert} \\
&\geq & \frac{\vert\sum_{k=1}^n z_k\vert}{1+\vert\sum_{k=1}^n z_k\vert}+\frac{\vert z_{n+1} \vert}{1+\vert z_{n+1} \vert} \\
&\geq & \frac{\vert\sum_{k=1}^n z_k\vert+\vert\ z_{n+1} \vert}{1+\vert\sum_{k=1}^n z_k\vert +\vert\ z_{n+1} \vert} \\
\end{eqnarray*}
