\(\Rightarrow a^{2} + b^{2} – c^{2} = ab\)

\(\cos C = \displaystyle \frac{a^{2} + b^{2} – c^{2}}{2ab} = \displaystyle \frac{1}{2}\)

\(C \in (0, \pi) \Rightarrow C = \displaystyle \frac{\pi}{3}\)

\(\because \displaystyle \frac{\sin B\sin C}{3\sin A} = \displaystyle \frac{\cos A}{a} + \displaystyle \frac{\cos C}{c}\)

\(\therefore \displaystyle \frac{\displaystyle \frac{b}{2R}\sin C}{3\displaystyle \frac{a}{2R}} = \displaystyle \frac{c\cos A + a\cos C}{ac}\)

\(\displaystyle \frac{b\sin \displaystyle \frac{\pi}{3}}{3a} = \displaystyle \frac{c\cos A + a\cos C}{ac}\)

如图：\(A\)

\(c\cos A = AD\)，\(a\cos C = CD\)

\(\therefore c\cos A + a\cos C = AC = b\)

即原式为\(\displaystyle \frac{\displaystyle \frac{\sqrt{3}}{2}b}{3a} = \displaystyle \frac{b}{ac}\)

\(\Rightarrow \displaystyle \frac{\sqrt{3}}{6} = \displaystyle \frac{1}{c}\) \(\therefore c = 2\sqrt{3}\)

\(C\)为定角\(\displaystyle \frac{\pi}{3}\)

\(c\)为定弦\(2\sqrt{3} \Rightarrow\)外接圆

如图，\(\triangle ABC\)为等边\(\triangle\)时\((a + b)_{\max} = 4\sqrt{3}\)

\(BC\)为\(2R\)即\(\triangle ABC\)为\(Rt\triangle\)时\((a + b)_{\min} = 6\)

\(\therefore a + b \in (6, 4\sqrt{3}]\)

当\(a + b\)最大时\(\displaystyle \frac{c}{a + b}\)最小

\(\therefore \displaystyle \frac{c^{2}}{a + b} \in [\sqrt{3}, 2)\)