Want less brutish proof: if $a+b+c=3abc$ then $\frac1a+\frac1b+\frac1c\geq 3$

I will use that

$$a+b+c=3abc \Leftrightarrow \frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=3$$

By AM-GM $$\left(\frac{1}{a}+\frac{1}{b}\right)^2\ge\frac{4}{ab}\\ \left(\frac{1}{a}+\frac{1}{c}\right)^2\ge\frac{4}{ac}\\ \left(\frac{1}{b}+\frac{1}{c}\right)^2\ge\frac{4}{bc}$$

so

$$\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2\ge 4\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=12\quad (1)$$

but $x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz)$, so

$$\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+9\right)\quad (2)$$

and also,

$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{ab}+\frac{1}{ab}+\frac{1}{bc}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-6\quad (3)$$

Now use $(3)$ in $(2)$ and get:

$$\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-6 \quad (4)$$

Now put $(4)$ in $(1)$:

$$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-6\ge 12\to \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3$$

We need to prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq3\sqrt{\frac{a+b+c}{3abc}}$$ or $$ab+ac+bc\geq\sqrt{3abc(a+b+c)}$$ or $$\sum_{cyc}(a^2b^2-a^2bc)\geq0$$ or $$\sum_{cyc}(a^2c^2+b^2c^2-2c^2ab)\geq0$$ or $$\sum_{cyc}c^2(a-b)^2\geq0.$$ Done!