An inequality involving complex numbers

Let $z_1, z_2, \dots, z_n$ be complex numbers . Prove that:

$$\left ( \sum_{k=1}^{n}\left | z_k \right | \right )^2 -\left | \sum_{k=1}^{n}z_k \right |^2 \geq \left ( \sum_{k=1}^{n}\left | \mathfrak{Re}\left ( z_k \right ) \right | - \left | \sum_{k=1}^{n}\mathfrak{Re}\left ( z_k \right ) \right | \right )^2$$

(G. Stoika, Canada)

Solution  (by Roberto Tauraso/ University of Rome/Italy)

 Let $z_k = x_k +i y_k , \; k=1, \dots, n$ then by Cauchy - Schwartz inequality it holds that:

$$\left | z_j \right |\left | z_k \right |= \sqrt{\left | x_j \right |^2 + \left | y_j \right |^2} \cdot \sqrt{\left | x_k \right |^2+ \left | y_k \right |^2}\geq \left | x_j \right |\left | x_k \right |+ \left | y_j \right |\left | y_k \right |\geq \left | x_j \right |\left | x_k \right |+ y_j y_k$$

Thus:

\begin{align*}
\left ( \sum_{k=1}^{n}\left | z_k \right | \right )^2 -\left | \sum_{k=1}^{n}z_k \right |^2 &=2 \sum_{1 \leq j<k \leq n}\left | z_j \right | \left | z_k \right | - 2 \sum_{1 \leq j <k \leq n}\mathfrak{Re}\left ( \overline{z_j}z_k \right ) \\
 &= 2 \sum_{1\leq j < k \leq n} \left ( \left | z_j \right | \left | z_k \right | - (x_j x_k +y_j y_k) \right )\\
 &\geq 2 \sum_{1\leq j <k \leq n}\left ( \left | x_j  \right |\left | x_k \right | - x_j x_k \right )\\
 &= \left ( \sum_{k=1}^{n}\left | x_k \right | \right )^2 - \left | \sum_{k=1}^{n}x_k \right |^2\\
 &= \left ( \sum_{k=1}^{n}\left | x_k \right | + \left | \sum_{k=1}^{n} x_k \right | \right )\left ( \sum_{k=1}^{n}\left | x_k \right | - \left | \sum_{k=1}^{n}x_k \right | \right ) \\
 &\geq \left ( \sum_{k=1}^{n}\left | x_k \right | - \left | \sum_{k=1}^{n}x_k \right | \right )^2 \\
 &=\left ( \sum_{k=1}^{n}\left | \mathfrak{Re}(z_k) \right | - \left | \sum_{k=1}^{n}\mathfrak{Re}(z_k) \right | \right )^2
\end{align*}

proving the result. The well known identity $\sum \limits_{k=1}^{n}|x_k| \geq \left | \sum \limits_{k=1}^{n}x_k \right |$ was also used.