a

b

c

R

\frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{(c+a)(a+b)}\geqslant\frac{1}{4R^{2}}.

a+b+c\leqslant3\sqrt{3}R~\mbox{и}~\frac{(a+b)+(b+c)+(c+a)}{3}\geqslant\sqrt[{3}]{{(a+b)(b+c)(c+a)}}

\frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{(c+a)(a+b)}=

=\frac{(a+b)+(b+c)+(c+a)}{(a+b)(b+c)(c+a)}=\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}\geqslant

\geqslant\frac{2(a+b+c)}{\frac{8}{27}(a+b+c)^{3}}=\frac{27}{4(a+b+c)^{2}}\geqslant\frac{27}{4\cdot27R^{2}}=\frac{1}{4R^{2}}.