Prove that $\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} \ge \frac{{55}}{{12\left( {a + b + c} \right)}}$ 
Let $a,b,c>0,\left( {a + b + c} \right)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) =10.$ Prove that
$$\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} \ge \frac{{55}}{{12\left( {a + b + c} \right)}}.$$

Side Note: The familiar inequality is:
$\frac{1}{{a + b}} + \frac{1}{{b + c}} + \frac{1}{{c + a}} \ge \frac{{9}}{{2\left( {a + b + c} \right)}}.$

It suffices to prove that
$$\frac{a}{b + c}  + \frac{b}{c + a} + \frac{c}{a + b} \ge \frac{19}{12}.$$
Since the condition and the inequality are both homogeneous and symmetric, assume that $c = \max(a, b, c) = 1$.
Let $p = a + b, q = ab$. We have $0 < p \le 2$.
It suffices to prove that
$$\frac{p^2 - 2q + p}{1 + p + q} + \frac{1}{p} \ge \frac{19}{12}.$$
From $(a + b + c)(1/a + 1/b + 1/c) = 10$, we have $(p + 1)(p/q + 1) = 10$
and $q = p(1 + p)/(9 - p)$.
It suffices to prove that
$$\frac{p^2 - 2\cdot p(1 + p)/(9 - p) + p}{1 + p + p(1 + p)/(9 - p)} + \frac{1}{p} \ge \frac{19}{12}$$
or
$$\frac{(4 - p)(3 - 2p)^2}{36p} \ge 0 $$
which is true.
We are done.
