57. (2007.01.29) Proof: We will prove a stronger form of the inequality:
$$\sum \frac{a(b+c)}{a^{2}+b c} \geqslant 2+\frac{8 a^{2} b^{2} c^{2}}{\prod\left(a^{2}+b c\right)}$$

When one of $a, b, c$ is zero, assume without loss of generality that $c=0$, then inequality (1) becomes
$$\frac{b}{a}+\frac{a}{b} \geqslant 2$$

which is clearly true.
When $a, b, c$ are all non-zero, let $\frac{b}{a}=x, \frac{c}{b}=y, \frac{a}{c}=z, x, y, z \in \overline{\mathbf{R}^{-}}, x y z=1$, then inequality (1) becomes
$$\begin{array}{l} \sum \frac{1+y z}{y+z} \geqslant 2+\frac{8}{\prod(y+z)} \Leftrightarrow \\ \sum(1+y z)(x+y)(x+z) \geqslant 2 \prod(y+z)+8 \Leftrightarrow \\ \sum x^{2}+3 \sum y z+\sum x+\left(\sum y z\right)^{2} \geqslant 2 \sum x \cdot \sum y z+6 \Leftrightarrow \\ \left(\sum x\right)^{2}+\left(\sum y z\right)^{2}+\sum(y z+x) \geqslant 2 \sum x \cdot \sum y z+6 \end{array}$$

Since
so
and
$$\begin{array}{c} x y z=1 \\ \sum(y z+x) \geqslant 6 \\ \left(\sum x\right)^{2}+\left(\sum y z\right)^{2} \geqslant 2 \sum x \cdot \sum y z \end{array}$$

Therefore, inequality (3) holds, which means inequality (2) holds, and thus inequality (1) is proved.