We homogenize the inequality by multiplying the LHS by $abc$ to get \[\sum_{\text{cyc}}\frac{a}{b^2(a+c)(b+c)}=\sum_{\text{cyc}}\frac{a^2bc}{b^2(a+c)(b+c)}.\]Multiply top and bottom by $\frac{c}{b}$ to get \[\sum_{\text{cyc}}\frac{a^2bc}{b^2(a+c)(b+c)}=\sum_{\text{cyc}}\frac{a^2c^2}{bc(a+c)(b+c)}.\]By Titu's Lemma we have that \[\sum_{\text{cyc}}\frac{a^2c^2}{bc(a+c)(b+c)}\ge\frac{\left(\sum_{\text{cyc}}ac\right)^2}{\sum_{\text{cyc}}bc(a+c)(b+c)}.\]Expand to get \[\frac{\left(\sum_{\text{cyc}}ac\right)^2}{\sum_{\text{cyc}}bc(a+c)(b+c)}=\frac{2\sum_{\text{cyc}}a^2bc+\sum_{\text{cyc}}a^2b^2}{3\sum_{\text{cyc}}a^2bc+\sum_{\text{cyc}}a^2b^2}.\]As we WTS \[\frac{2\sum_{\text{cyc}}a^2bc+\sum_{\text{cyc}}a^2b^2}{3\sum_{\text{cyc}}a^2bc+\sum_{\text{cyc}}a^2b^2} \ge \frac{3}{4},\]problem reduces to showing that \[\sum_{\text{cyc}}a^2b^2\ge\sum_{\text{cyc}}a^2bc\]which is true by Miurhead's.