Can't apply Titu's directly

$\begin{aligned} \displaystyle \sum_{\text{cyc}} \dfrac{ab+bc+ca}{a(a+2b)} &= \displaystyle \sum_{\text{cyc}} \dfrac{a(b+c)+bc}{a(a+2b)}\\ &= \displaystyle \sum_{\text{cyc}} \dfrac{b+c}{a+2b} + \displaystyle \sum_{\text{cyc}} \dfrac{bc}{a(a+2b)} \end{aligned}$

We will apply Cauchy-Schwarz on both these cyclic summations.

$\begin{aligned} \left ( \displaystyle \sum_{\text{cyc}} \dfrac{b+c}{a+2b} \right ) \left ( \displaystyle \sum_{\text{cyc}} (b+c)(a+2b) \right ) &\geq \left ( \displaystyle \sum_{\text{cyc}} (b+c) \right )^2\\ \displaystyle \sum_{\text{cyc}} \dfrac{b+c}{a+2b} &\geq \dfrac{\left ( \displaystyle \sum_{\text{cyc}} (b+c) \right )^2}{\displaystyle \sum_{\text{cyc}} (b+c)(a+2b)}\\ &= \dfrac {4 ( a+b+c)^2}{2(a^2+b^2+c^2)+4(ab+bc+ca)}\\ &= 2\\ \end{aligned}$

$\begin{aligned} \left ( \displaystyle \sum_{\text{cyc}} \dfrac{bc}{a(a+2b)} \right ) \left ( \displaystyle \sum_{\text{cyc}} abc (a+2b) \right ) &\geq \left ( \displaystyle \sum_{\text{cyc}} bc \right )^2\\ \left ( \displaystyle \sum_{\text{cyc}} \dfrac{bc}{a(a+2b)} \right ) &\geq \dfrac {\left ( \displaystyle \sum_{\text{cyc}} bc \right )^2}{ abc \left ( \displaystyle \sum_{\text{cyc}} (a+2b) \right )}\\ &= \dfrac {\left ( \displaystyle \sum_{\text{cyc}} bc \right )^2}{ 3abc (a+b+c)}\\ &= 1 + \dfrac{\displaystyle \sum_{\text{cyc}} a^2 (b-c)^2}{6abc (a+b+c)}\\ &\geq 1\\ \end{aligned}$

Therefore, their sum must be greater than or equal to 3, with equality at $a=b=c$ .