Notice that
$$\frac{c}{a+2b} = \frac{a+2b+c}{a+2b} - 1, \quad \frac{a+2b}{c} = \frac{a+2b+c}{c} - 1$$
So we have:
$$\begin{aligned} \sum_{cyc} \frac{c}{a+2b} + \sum_{cyc} \frac{a+2b}{c} &= \sum_{cyc} (a+2b+c) \left( \frac{1}{a+2b} + \frac{1}{c} \right) - 8 \\ &= \sum_{cyc} \frac{(a+2b+c)^2}{c(a+2b)} - 8. \end{aligned}$$
Using the Cauchy-Schwarz inequality:
$$\left( \sum_{cyc} c(a+2b) \right) \left( \sum_{cyc} \frac{(a+2b+c)^2}{c(a+2b)} \right) \geq 16(a+b+c+d)^2$$
that is
$$\begin{aligned} \sum_{cyc} \frac{(a+2b+c)^2}{c(a+2b)} - 8 &\geq 8 \left( \frac{(a+b+c+d)^2}{\sum_{cyc} c(a+2b)} - 1 \right) \\ &= 8 \left( \frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd} - 1 \right). \end{aligned}$$