$\sum_{cyc} \frac{c}{a+2b}+\sum_{cyc} \frac{a+2b}{c}+8=\sum_{cyc}\frac{a+2b+c}{a+2b}+\sum_{cyc} \frac{a+2b+c}{c}$ $=\sum_{cyc} \frac{(a+2b+c)^2}{(a+2b)c} \ge \frac{(\sum_{cyc} (a+2b+c))^2}{\sum_{cyc} (a+2b)c}=8\frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd}$

Solved with hint Note that $$\frac{c}{a+2b}+\frac{a+2b}{c}+2=\frac{(a+2b+c)^2}{(a+2b)c},$$so we wish to prove that $$\sum_{\mathrm{cyc}} \frac{(a+2b+c)^2}{(a+2b)c} \geq \frac{8(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd}.$$This follows by Cauchy-Schwarz: we wish to prove that $$(4a+4b+4c+4d)^2 \geq 16(a+b+c+d)^2,$$which is obvious. $\blacksquare$