Question 146
Given $a, b, c, d \gt 0$, prove that
$$
\sum_{cyc} \frac{c}{a + 2b} + \sum_{cyc} \frac{a + 2b}{c} \geq 8 \left( \frac{(a + b + c + d)^2}{ab + ac + ad + bc + bd + cd} - 1 \right),
$$
where $\sum_{cyc} f(a, b, c, d) = f(a, b, c, d) + f(d, a, b, c) + f(c, d, a, b) + f(b, c, d, a)$.
$$
\sum_{cyc} \frac{c}{a + 2b} + \sum_{cyc} \frac{a + 2b}{c} \geq 8 \left( \frac{(a + b + c + d)^2}{ab + ac + ad + bc + bd + cd} - 1 \right),
$$
where $\sum_{cyc} f(a, b, c, d) = f(a, b, c, d) + f(d, a, b, c) + f(c, d, a, b) + f(b, c, d, a)$.