58. If $a, b, c$ are all positive real numbers, and $a+b+c=1$, find the largest real number $M$ such that the inequality $(a+b c)(b+c a)(c+a b)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant M a b c$. (2011 Uzbekistan Mathematics
58. Since $a+b+c=1$, we have $a+bc=a(a+b+c)+bc=(a+b)(c+a)$, so $(a+bc)(b+ca)(c+ab)=(a+b)^{2}(b+c)^{2}(c+a)^{2}$. The problem is reduced to finding the largest real number $M$ such that the inequality
$$(a+b)^{2}(b+c)^{2}(c+a)^{2}(ab+bc+ca) \geqslant M(abc)^{2}(a+b+c)^{2}$$

holds. Given that $(a+b)(b+c)(c+a) \geqslant \frac{8}{9}(a+b+c)(ab+bc+ca)$ and $ab+bc+ca \geqslant 3 \sqrt[3]{(abc)^{2}}$, we have $(a+b)^{2}(b+c)^{2}(c+a)^{2}(ab+bc+ca) \geqslant \frac{64}{81}(a+b+c)^{2}(ab+bc+ca)^{3} \geqslant \frac{64}{81}(a+b+c)^{2} \times 27(abc)^{2}=\frac{64}{3}(a+b+c)^{2}(abc)^{2}$. Therefore, the largest real number $M=\frac{64}{3}$.