Solution:
By the Cauchy-Schwarz Inequality, we have
$$\frac{1}{a+b+c}+\frac{1}{a+c} \geqslant \frac{4}{2 a+b+2 c}$$
and
$$\frac{1}{b+c}+\frac{1}{a+b} \geqslant \frac{4}{a+2 b+c}$$
Since
$$a^{2}+a c+c^{2}=\frac{3}{4}(a+c)^{2}+\frac{1}{4}(a-c)^{2} \geqslant \frac{3}{4}(a+c)^{2}$$
then, writing $L$ for the Left Hand Side of the required inequality, we get
$$L \geqslant \frac{3(a+c)^{2}}{2 a+b+2 c}+\frac{4 b^{2}}{a+2 b+c}$$
Using again the Cauchy-Schwarz Inequality, we have:
$$L \geqslant \frac{(\sqrt{3}(a+c)+2 b)^{2}}{3 a+3 b+3 c}>\frac{(\sqrt{3}(a+c)+\sqrt{3} b)^{2}}{3 a+3 b+3 c}=a+b+c$$

Alternative Solution by PSC:
The required inequality is equivalent to
$$\left[\frac{b^{2}}{a+b}-(b-a)\right]+\frac{b^{2}}{b+c}+\left[\frac{a^{2}+a c+c^{2}}{a+c}-a\right]+\left[\frac{a^{2}+a c+c^{2}}{a+b+c}-(a+c)\right]>0$$
or equivalently, to
$$\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}>\frac{a b+b c+c a}{a+b+c}$$
However, by the Cauchy-Schwarz Inequality we have
$$\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a} \geqslant \frac{(a+b+c)^{2}}{2(a+b+c)} \geqslant \frac{3(a b+b c+c a)}{2(a+b+c)}>\frac{a b+b c+c a}{a+b+c}$$