Question 93
Show that for positive real numbers $a$, $b$, and $c$,
$$
\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \geq \frac{(a+b+c)^{2}}{a b(a+b)+b c(b+c)+c a(c+a)}
$$
$$
\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \geq \frac{(a+b+c)^{2}}{a b(a+b)+b c(b+c)+c a(c+a)}
$$