Since $$\prod (ab+2(a+b)-9)=\prod\left(\frac{1}{c}+2(5-c)-9\right)=\prod (1-c)\left(2+\frac{1}{c}\right)$$hence with $a,b,c>0$ we must have $$(1-a)(1-b)(1-c)\ge 0$$or equivalently $$ab+bc+ca\ge 5$$and thus $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}\ge 5$$The minimum is attained for any permutation of $(2-\sqrt{3},1,2+\sqrt{3})$.

Let $a,b,c$ be positive reals such that $abc=1$, $a+b+c=4$ and $(ab+2a+2b-7)(bc+2b+2c-7)(ca+2c+2a-7)\geq 0$. Prove that $$\frac {1}{a}+ \frac {1}{b}+ \frac{1}{c}\geq 4$$Let $a,b,c$ be positive reals such that $abc=1$, $a+b+c=4$ and $(ab+a+b-4)(bc+b+c-4)(ca+c+a-4)\geq 0$. Prove that$$\frac {1}{a}+ \frac {1}{b}+ \frac{1}{c}\geq 4$$

The original problem is a special case of the following generalization where $$\lambda=1,\beta=1,\theta=2,p=5$$. Minimum value 5 is attined by giving these value.

Generalization 1 Let $a,b,c$ be positive reals suc that $abc=1$ , $a+b+c=p$ and $$\left(\lambda ab+\theta a+\theta b-\theta p +\beta\right)\left(\lambda bc+\theta b+\theta c-\theta p +\beta\right)\left(\lambda ca+\theta c+\theta a-\theta p +\beta\right)\geq 0$$holds. Then prove that $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq \dfrac{p\left(\beta+\sqrt{\beta^2+4\theta\lambda}\right)}{2\theta}+\dfrac{2\theta}{\beta+\sqrt{\beta^2+4\theta\lambda}}-\left(\dfrac{\beta+\sqrt{\beta^2+4\theta\lambda}}{2\theta}\right)^2$$