What is the least value of the expression? If p , q , r are all positive numbers.
And if


p + q + r =1


then what is the least value of 
$$\left(\frac{1-p}{p}\right) \left(\frac{1-q}{q}\right) \left(\frac{1-r}{r}\right)  $$ 
?
I begin with keeping 1-p = q + r
So the expression becomes
$$\left(\frac{q+r}{p}\right) \left(\frac{p+r}{q}\right) \left(\frac{p+q}{r}\right)  $$ 
And then i try to use A.M. >= G.M inequality.
But cant get to the answer.
Thanks in advance.
 A: $$(1-p)(1-q)(1-r) = 1- (p+q+r)+(pq+qr+rp) -pqr = (pq+qr+rp) -pqr$$
Hence,$$ \frac{(1-p)(1-q)(1-r)}{pqr} = \frac1r + \frac1p + \frac1q - 1$$
Now make use of the arithmetic mean-harmonic mean inequality (or equivalently arithmetic mean-geometric mean inequality) to get $$\frac1r + \frac1p + \frac1q \geq 9 \left(p+q+r\right) = 9$$
Hence, $$\frac{(1-p)(1-q)(1-r)}{pqr} \geq 8$$ Equality holds when $p=q=r=\frac13$
A: Given, $p+q+r=1$ and we have to find the minimum value of $\bigg(\dfrac{1-p}{p}\bigg)\bigg(\dfrac{1-q}{q}\bigg)\bigg(\dfrac{1-r}{r}\bigg)$, we proceed just as you did.  
$\bigg(\dfrac{1-p}{p}\bigg)\bigg(\dfrac{1-q}{q}\bigg)\bigg(\dfrac{1-r}{r}\bigg)=\bigg(\dfrac{q+r}{p}\bigg) \bigg(\dfrac{p+r}{q}\bigg) \bigg(\dfrac{p+q}{r}\bigg)=\bigg(\dfrac{q+r}{q}\bigg) \bigg(\dfrac{p+r}{r}\bigg) \bigg(\dfrac{p+q}{p}\bigg)=\bigg(1+\dfrac{r}{q}\bigg) \bigg(1+\dfrac{p}{r}\bigg) \bigg(1+\dfrac{q}{p}\bigg)$  
Now, by AM-GM inequality,  
$\bigg(1+\dfrac{r}{q}\bigg) \bigg(1+\dfrac{p}{r}\bigg) \bigg(1+\dfrac{q}{p}\bigg)\ge 2\cdot\sqrt{\dfrac{r}{q}}\cdot2\sqrt{\dfrac{p}{r}}\cdot2\sqrt{\dfrac{q}{p}}=8\cdot\sqrt{\dfrac{pqr}{pqr}}=8$  
I think you can find out yourself when equality occurs.
