48. (2007.04.04) Simplify and prove: From $\sum a^{2}=3$, we know that the original inequality is equivalent to
$$\begin{array}{l} 3 \sum a^{2} \cdot \sum b c-\left(\sum a^{2}\right)^{2} \geqslant 6 a b c \cdot \sqrt{3 \sum a^{2}} \Leftrightarrow \\ \sqrt{3 \sum a^{2}}\left(3 \sum b c-\sum a^{2}\right) \geqslant 18 a b c \end{array}$$

Since
$$\sqrt{3 \sum a^{2}} \geqslant \sum a$$

It suffices to prove
$$\sum a\left(3 \sum b c-\sum a^{2}\right) \geqslant 18 a b c$$

Let the circumradius of $\triangle ABC$ be $R$, the inradius be $r$, and the semiperimeter be $s$, then
$$\begin{array}{c} a b c=2 R r \sum a \\ 3 \sum b c-\sum a^{2}=\frac{1}{4}\left(\sum a\right)^{2}+\frac{5}{4}\left(2 \sum b c-\sum a^{2}\right)= \\ s^{2}+5 r(4 R+r) \end{array}$$

Thus, to prove inequality (1), it suffices to prove
$$s^{2}+5 r(4 R+r) \geqslant 36 R r \Leftrightarrow s^{2} \geqslant 16 R r-r^{2}$$

This inequality is the Gerretsen inequality, hence the original inequality is proved.