If $a,b$ and $c$ are positive real numbers such that $a+b+c=1$, what is the minimum value of $(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$ ?
How can we solve this? I tried to simplify the expression but still can't figure a way to solve this.
My simplified expression : $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1$
$$\begin{align} (\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1) &=\frac{1}{abc}-\frac{1}{ab}-\frac{1}{bc}-\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1 \\ &=\frac{1-a-b-c}{abc}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1 \\ &=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1 \\ &\geq\frac{3}{\frac{a+b+c}{3}}-1 & \text{by AM-HM}\\ &=9-1 \\ &=8 \end{align}$$
with equality if and only if $a=b=c$.
$$\frac{1}{a}-1=\frac{1-a}{a}=\frac{b+c}{a}=\frac{b}{a}+\frac{c}{a}$$
Similarly,
$$\frac{1}{b}-1=\frac{a}{b}+\frac{c}{b}$$
$$\frac{1}{c}-1=\frac{a}{c}+\frac{b}{c}$$
$\therefore \left(\frac{1}{a}-1 \right)\left(\frac{1}{b}-1 \right)\left(\frac{1}{c}-1 \right)$ $=\left(\frac{b}{a}+\frac{c}{a}\right)\left(\frac{a}{b}+\frac{c}{b}\right)\left(\frac{a}{c}+\frac{b}{c}\right)$
$\ge \left(2 \sqrt{\frac{bc}{a^2}} \right) \left(2 \sqrt{\frac{ca}{b^2}} \right) \left(2 \sqrt{\frac{ab}{c^2}} \right)$
$=8$
Equality holds
iff $\frac{b}{a}=\frac{c}{a},\frac{a}{b}=\frac{c}{b} $ and $\frac{a}{c}=\frac{b}{c}$
iff $a=b=c$.
The answers posted all uses QM-AM-GM-HM Inequality, so this is an alternative Cauchy-Schwarz proof by making use of the fact that $a+b+c=1$.
As OP has pointed out, the expression $(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$ simplifies to $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1$.
$$\begin{align} \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1 &=(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(a+b+c)-1 \\ &\geq(1+1+1)^2-1 &\text{by Cauchy-Schwarz Inequality}\\ &=9-1 \\ &=8 \end{align}$$
Equality is achieved if and only if $\frac{\frac{1}{a}}{a}=\frac{\frac{1}{b}}{b}=\frac{\frac{1}{c}}{c}$, i.e. $a=b=c=\frac{1}{3}$.