For any positive real numbers $a,b,c$ then prove $$(a+b+c-3)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3\right)+abc+\frac{1}{abc}\ge 2.$$
I've try to assume $a+b+c\ge 3.\quad(1)$
By AM-GM $$abc+\frac{1}{abc}\ge 2$$ and we need to prove $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3\ge 0$$
By Cauchy-Schwarz $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}$$but $$\frac{9}{a+b+c}\ge 3 \iff a+b+c\le 3$$ which gives a contradiction to $(1).$
Similarly when $a+b+c<3$ I can not find a good approach for each assumed condition.
Could you please give some hint to solve it? Thank you.
Best Answer
Since among three numbers: $$(a-1)(bc-1), (b-1)(ca-1), (c-1)(ab-1)$$there are at least two of them having the same sign. WLOG, assume that they are $(b-1)(ca-1)$, and $(c-1)(ab-1)$.
We have: $$ \left(a+b+c -3\right) \left(\frac{1}{a} +\frac{1}{b} +\frac{1}{c} -3\right) +abc +\frac{1}{abc} -2$$
$$= \left( a+\frac{1}{a}-2\right) \left(b+\frac{1}{b} +c+\frac{1}{c}-4 \right)+\frac{(b-1)(c-1)(ab-1)(ac-1)}{abc} \ge 0.$$