# 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.

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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 $p=q=r=\frac13$
This is a full answer. The question is tagged (homework)... –  user2468 Mar 7 '12 at 5:43
@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