My classmate told me this problem yesterday. The problem is that: let $a,b,c,d>0$, prove that: $\sum_{cyc}{\frac{c}{a+2b}}+\sum_{cyc}\frac{c+2d}{a}\geq8(\frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd}-1)$
I tried to expand it and used Cauchy-Schwarz-inequality,then I got the following result: $$ \begin{aligned}\sum_{cyc}{\frac{c}{a+2b}}+\sum_{cyc}\frac{c+2d}{a}&={\frac{c}{a+2b}}+{\frac{d}{b+2c}}+{\frac{a}{c+2d}}+{\frac{b}{d+2a}}+\frac{c+2d}{a}+\frac{d+2a}{b}+\frac{a+2b}{c}+\frac{b+2c}{d}\\&\geq\frac{[4(a+b+c+d)]^2}{4(ab+ac+ad+bc+bd+cd)}\\&=4(\frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd})\end{aligned} $$ BUT this doesn't look right.
Note that your inequality can be rearranged to \begin{eqnarray*} \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}} \\ & \geq & 8\frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd} \end{eqnarray*} Note also that \begin{eqnarray*} \sum_{cyc}(a+2b)c =2(ab+ac+ad+bc+bd+cd). \end{eqnarray*} Multiply through by this & now apply CS.