Let $a,b,c \in \mathbb R^+$ and $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} =1$$
Show that $$(a^2 -3a +3)(b^2-3b+3)(c^2-3c+3) \ge 27$$
I tried using using the AM-GM inequality and some algebraic manipulation to who each of the quadratic terms cannot be smaller than $0.75$ and other little results but I am struggling to be able to solve this. Any help would be appreciated.
For positive variables let $\frac{1}{a}=\frac{x}{3},$ $\frac{1}{b}=\frac{y}{3}$ and $\frac{1} {c}=\frac{z}{3}.$
Thus, $x+y+z=3$ and we need to prove that $$\sum_{cyc}\left(\ln\left(\frac{9}{x^2}-\frac{9}{x}+3\right)-\ln3\right)\geq0$$ or $$\sum_{cyc}\left(\ln(x^2-3x+3)-2\ln{x}\right)\geq0$$ or $$\sum_{cyc}\left(\ln(x^2-3x+3)-2\ln{x}+3(x-1)\right)\geq0.$$ Let $f(x)=\ln(x^2-3x+3)-2\ln{x}+3(x-1).$
Thus, $$f'(x)=\frac{2x-3}{x^2-3x+3}-\frac{2}{x}+3=\frac{3(x-1)(x^2-2x+2)}{x(x^2-3x+3)},$$ which gives $x_{min}=1$ and we are done!