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- Feb 14, 2012

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I have encountered a problem and I am not being able to figure out the answer.

Problem:

Given that $p, q, r, s$ are all positive real numbers and they satisfy the system

$p+q+r+s=12$

$pqrs=27+pq+pr+ps+qr+qs+rs$

Determine $p, q, r$ and $s$.

Attempt:

The AM-GM inequality for both $p, q, r, s$ and $pq,pr,ps,qr,qs,rs$ are:

1. | $\dfrac{p+q+r+s}{4} \ge \sqrt[4]{pqrs}$ which then gives $(\dfrac{12}{4})^4 \ge pqrs$ or $pqrs \le 81$ |

2. | $\dfrac{pq+pr+ps+qr+qs+rs}{6} \ge \sqrt[6]{(pqrs)^3}$ which then gives $(\dfrac{pqrs-27}{6})^2 \ge pqrs$ $(pqrs-81)(pqrs-81) \ge 0$ $pqrs \le 9$ or $pqrs \ge 81$ |

After that, I don't see how to proceed...should I conclude that since we need to find $pqrs$ that satisfy both of the inequalities below

$pqrs \le 81$ and $pqrs \ge 81$

$\therefore pqrs=81$ and and obviously the answer would be $p=q=r=s=3$?