Write $x = \log a$, $y = \log b$ and $z = \log c$, $x, y, z > 0$. Your statement comes to proving $\frac{x + y}{z} + \frac{y + z}{x} + \frac{z + x}{y} \ge 4\left( \frac{x}{y + z} + \frac{y}{z + x} + \frac{z}{x + y} \right)$. However, this follows from $\frac{x}{y} + \frac{x}{z} \ge \frac{4x}{y + z} \Leftrightarrow (y - z)^2 \ge 0$ and the similar relations.

let $log_ab=\frac{x}{y},log_bc=\frac{y}{z},log_ca=\frac{z}{x}$ it's equivalent to $\sum(\frac{x}{y}+\frac{y}{x})\ge 4\sum\frac{y}{x+z}$ by expansion $\sum_{sym}x^4y^2-2\sum_{cyc}x^4yz+2\sum_{cyc}x^3y^3-6x^2y^2z^2\ge 0$ trivially true by Muirhead