Let $x, y, z$ denote $\frac1a, \frac1b, \frac1c,$ respectively, which are clearly positive reals. Then, the problem wants us to show that that if $xy+yz+zx \ge 1$, then at least $2$ of $2x + 3y + 6z \ge 6, 2y+3z+6x \ge 6,$ and $2z + 3x + 6y \ge 6$ are true. We'll prove this by showing its contrapositive. Suppose that $2x+3y+6z < 6, 2y+3z+6x < 6$ are true, then we claim that $xy+yz+zx < 1.$ The problem would then following by proving two symmetric statements. To show this, it suffices by scaling to show that when $2x+3y+6z = 6, 2y+3z+6z \le 6,$ then we have $xy+yz+zx \le 1.$ Then, let $t = 2y+3z+6x \le 6$. We can now easily solve for $y, z$ in terms of $x, y$: $$y = (2t-6) - 10x, z = (4-t) + \frac{14}{3}x.$$Then, we have that: $$xy+yz+zx = -52x^2 + (\frac{61}{3}t-70)x + (2t-6)(4-t),$$ and so since $t = 2y+3z + 6x \ge \frac12 (2x+3y+6z) \ge 3 \Rightarrow \frac{61}{3}t > 70$, we have that this is maximal when $x = \frac{\frac{61}{3}t - 70}{104},$ which we know is positive by $\frac{61}{3}t - 70 > 0.$ Plugging this in for $x$, we have that: $$xy + yz + zx \le -52 (\frac{\frac{61}{3}t - 70}{104})^2 + (2t-6)(4t-6).$$This implies that: $$208(xy+yz+zx) \le -(\frac{61}{3}t - 70)^2 + 208(8t^2 - 36t+36).$$ This is an upwards parabola with a vertex at $\frac{20886}{11255} < 2$, and so hence is maximized when $t = 6$. When $t = 6$, we know that $x = \frac{\frac{61}{3}t - 70}{104} = \frac{122 - 70}{104} = \frac12$, and so plugging $x, t$ into our equations for $y, z$ give that $y = 1, z = \frac13.$ In this case, $xy + yz + zx = 1$, and so since this was our maximal case, we are done. $\square$