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.

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 $$8 \le \frac{1}{a}+\frac {1}{b}+\frac {1}{c}-1$$
