Homogenizing an inequality with a condition

Question

I am stuck on the following problem, proving this with the condition that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$: $$(a+1)(b+1)(c+1)\geq 64$$ I started by expanding the lhs, $abc+ab+bc+ac+a+b+c+1$, but now I do not know what to do. I have also found that $\frac{ab+bc+ac}{abc}=1\Rightarrow abc=ab+bc+ac$, but I don't think this helps much. Would anyone be kind enough to provide some hints as to how I can homogenize this thing?
P.S. I am aware of some of the other "easier" solutions, like the one involving the AM-GM-HM inequality, but here I really just want to practice on how to homogenize inequalities.

Answer 1

$(a+1)(b+1)(c+1)\ge 64\Leftrightarrow 2(ab+bc+ca)+(a+b+c)\cdot \frac{abc}{ab+bc+ca}\ge 64\cdot \frac{a^2b^2c^2}{(ab+bc+ca)^2}$

This is what you want.