Solution:
Expanding and rearranging the denominator gives
$$a^{2} b + a b^{2} + b^{2} c + b c^{2} + c^{2} a + a c = c^{2}(a+b) + a^{2}(b+c) + b^{2}(c+a).$$
By Cauchy-Schwarz,
$$\left(c^{2}(a+b) + a^{2}(b+c) + b^{2}(c+a)\right)\left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}\right) \geq (a+b+c)^{2},$$
and dividing by $c^{2}(a+b) + a^{2}(b+c) + b^{2}(c+a)$ gives the desired inequality.