<s>[INST] If $a,b,c > 0$ and $a+b+c = 1$, then $(\tfrac{1}{a} −1)(\tfrac{1}{b} −1)(\tfrac{1}{c} −1) \geq 8$ 
If $a, b, c$ are positive real numbers and $a+b+c = 1$, prove that
  $$\left(\frac{1}{a} −1\right)\left(\frac{1}{b} −1\right)\left(\frac{1}{c} −1\right) \geq 8.$$

Thank you.
 [/INST] This following is my new methods:
let $$x=1-\dfrac{1}{a},y=1-\dfrac{1}{b},z=1-\dfrac{1}{c}$$
then
$$a+b+c=1\Longrightarrow \dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}=1$$
$$\dfrac{x}{x+1}=
\dfrac{1}{1+y}+\dfrac{1}{1+y}\ge2\sqrt{\dfrac{1}{(1+y)(1+z)}}\cdots (1)$$
$$\dfrac{y}{1+y}\ge2\sqrt{\dfrac{1}{(1+x)(1+z)}}\cdots (2)$$
$$\dfrac{z}{1+z}\ge 2\sqrt{\dfrac{1}{(1+x)(1+y)}}\cdots (3)$$
$(1)\times (2)\times (3)$
$$xyz\ge 8$$
I very like this methods, this methods can solve some hard inequaity,such as 
USA USAMO 1998
Problem
 http://www.artofproblemsolving.com/Forum/viewtopic.php?p=343867&sid=f8230676af97b2afacd78d0ccae6ad68#p343867
</s>