(2) $\frac{a^{5}+b^{5}}{a b(a+b)}=\frac{a^{4}+b^{4}-a b\left(a^{2}+b^{2}\right)+a^{2} b^{2}}{a+b}=$
$\frac{(a-b)^{4}+4 a b\left(a^{2}+b^{2}\right)-6 a^{2} b^{2}-a b\left(a^{2}+b^{2}\right)+a^{2} b^{2}}{a b} \geqslant$
$$\frac{3 a b\left(a^{2}+b^{2}\right)-5 a^{2} b^{2}}{a b}=3\left(a^{2}+b^{2}\right)-5 a b$$

Similarly, $\frac{b^{5}+c^{5}}{b c(b+c)} \geqslant 3\left(b^{2}+c^{2}\right)-5 b c, \frac{c^{5}+a^{5}}{c a(c+a)} \geqslant 3\left(c^{2}+a^{2}\right)-5 c a$, adding them up we get
$$\frac{a^{5}+b^{5}}{a b(a+b)}+\frac{b^{5}+c^{5}}{b c(b+c)}+\frac{c^{5}+a^{5}}{c a(c+a)} \geqslant 6-5(a b+b c+c a)$$

Since $a^{2}+b^{2}+c^{2}=1 \geqslant a b+b c+c a$, we have $6-5(a b+b c+c a) \geqslant 3(a b+$ $b c+c a)-2$. Therefore, (2) is a generalization of (1).