58. Simplified Proof

Original expression $\Leftrightarrow\left(\sum|a|+\left|\sum a\right|\right)^{2} \geqslant\left(\sum|b+c|\right)^{2} \Leftrightarrow$

However,
$$\sum(|b| \cdot|c|)+\sum\left(|a| \cdot\left|\sum a\right|\right) \geqslant \sum(|a+b| \cdot|a+c|)$$
$$\begin{array}{l} \sum(|b| \cdot|c|)+\sum\left(|a| \cdot\left|\sum a\right|\right)= \\ \sum(|b| \cdot|c|+|a| \cdot|a+b+c|) \geqslant \\ \sum\left|b c+a^{2}+a b+a c\right|=\sum|(a+b)(a+c)| \end{array}$$

Therefore, the original expression holds.