Cauchy-Schwarz inequality in $\mathbb{R}^{3}$ How to use Cauchy-Schwarz inequality in $\mathbb{R}^{3}$ to show that if $a+b+c=1$ then
$\displaystyle \left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right) \geq 8$?
Any idea how to take the vectors? Thanks a lot.
 A: With a little algebra, you can write $$\bigg{(}\frac{1}{a}-1\bigg{)}\bigg{(}\frac{1}{b}-1\bigg{)}\bigg{(}\frac{1}{c}-1\bigg{)} = \frac{1}{abc}(1-c-b+bc-a+ac+ab-abc) = \frac{1}{abc}(bc+ac+ab-abc) = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1 $$
Now, take $x = \bigg{(}\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}}\bigg{)}$ and $y=(\sqrt{a},\sqrt{b},\sqrt{c})$. Then $$\langle x,y \rangle^{2} = 3^{2} = 9 \le \bigg{(}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg{)}(a+b+c) = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
Thus $$8 \le \frac{1}{a}+\frac {1}{b}+\frac {1}{c}-1$$
A: Assuming $a,b,c >0$, we can write equivalently
$$
\displaystyle \left(1-a\right)\left(1-b\right)\left(1-c\right) \geq 8 a b  c
$$
Using the condition $a+b+c=1$, this is 
$$
\displaystyle \left(b+c\right)\left(a+c\right)\left(a+b\right) \geq 8 a b  c
$$
or
$$
\displaystyle \left(1+b/c\right)\left(1+c/a\right)\left(1+a/b\right) \geq 8 
$$
Expanding this gives
$$
\displaystyle \left(b/c+c/b\right)+\left(a/c+c/a\right)+\left(b/a+a/b\right) \geq 6
$$
The brackets are just there to guide the following argument, namely using
$$
\left(b/c+c/b\right) \ge 2 
$$
which is well known from Cauchy-Schwarz, see below. The same holds for the other two brackets, and hence the claim is proven.
Explanation: 
Cauchy-Schwarz gives that 
$$
(b + c) (1/b + 1/c) \ge 4
$$
Expanding this gives directly $
\left(b/c+c/b\right) \ge 2 
$.
This can also be obtained (if you don't want to use Cauchy-Schwarz) by just expanding the obviously correct inequality
$\left(\sqrt{\frac{b}{c}} - \sqrt{\frac{c}{b}}\right)^2 \ge 0 $
A: A slight alternative to @Andreas's answer is to note $a+b\ge 2\sqrt{ab}$ etc., so$$(1-c)(1-a)(1-b)=(a+b)(b+c)(c+a)\ge8abc.$$In particular, $a+b\ge 2\sqrt{ab}$ is a special case of Cauchy-Schwarz:$$a+b=\left|\binom{\sqrt{a}}{\sqrt{b}}\right|\left|\binom{\sqrt{b}}{\sqrt{a}}\right|\ge\binom{\sqrt{a}}{\sqrt{b}}\cdot\binom{\sqrt{b}}{\sqrt{a}}=2\sqrt{ab}.$$
