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Determine p, q, r and s

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  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Hi MHB,

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 $p=q=r=s=3$?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
On your second AM-GM inequality, you get $(pqrs-81)(pqrs-9) \ge 0$, with the individual inequalities that you found.

In the beginning of the problem, you specified that $p,q,r,s$ are real. Is that correct? If so, I see no way of nailing down all four values, given only two equations. Certainly, $p=q=r=s=3$ works, but what guarantee do we have that there isn't another solution? E.g., try setting $p=q=2$, and solving the resulting system for $r,s$, and see if there is a solution.

[EDIT] See Opalg's post below for a correction.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hi MHB,

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 $p=q=r=s=3$?
It looks as though you have solved this problem. You have shown that either $pqrs\leqslant9$ or $pqrs\geqslant 81$. But the equation $pqrs=27+pq+pr+ps+qr+qs+rs$ shows that $pqrs\geqslant27$, so that rules out the first of those possibilities. We are left with the second one, $pqrs\geqslant 81$. But you have also shown that $pqrs\leqslant 81$. Therefore $pqrs = 81$. That implies that equality occurs in the AM-GM inequality, and that only happens when all four quantities are equal. So $p=q=r=s=3$.
 
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  • #4

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
On your second AM-GM inequality, you get $(pqrs-81)(pqrs-9) \ge 0$, with the individual inequalities that you found.

In the beginning of the problem, you specified that $p,q,r,s$ are real. Is that correct? If so, I see no way of nailing down all four values, given only two equations. Certainly, $p=q=r=s=3$ works, but what guarantee do we have that there isn't another solution? E.g., try setting $p=q=2$, and solving the resulting system for $r,s$, and see if there is a solution.

[EDIT] See Opalg's post below for a correction.
Thanks Ackbach for your reply.

...We are left with the second one, $pqrs\geqslant 81$. But you have also shown that $pqrs\leqslant 81$. Therefore $pqrs = 81$. That implies that equality occurs in the AM-GM inequality, and that only happens when all four quantities are equal. So $p=q=r=s=3$.
Hi Opalg, thank you so much for pointing out that equality holds in the AM-GM inequality only if all of the quantities involved are equal...this is something I have totally forgotten about.:eek:

I understand it all now! Thanks guys!