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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.

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If this is homework, then what have you tried before? – chrisaycock Mar 7 '12 at 5:01

$$(1-p)(1-q)(1-r) = 1- (p+q+r)+(pq+qr+rp) -pqr = (pq+qr+rp) -pqr$$ 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 $p=q=r=\frac13$

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@Sivaram thanks.but i dint get how did you get this? $$\frac1r + \frac1p + \frac1q \geq 9 \left(p+q+r\right) = 9$$ – vikiiii Mar 8 '12 at 8:00
@vikiiii: It follows from Arithmetic mean Harmonic mean inequality (equivalently Arithmetic mean Geometric mean inequality). – user17762 Mar 9 '12 at 5:12

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.

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