How Can I Determine the Values of p, q, r, and s in This Mathematical Problem?

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In summary, the problem asks to determine the values of p, q, r, and s given that they are all positive real numbers and satisfy the system of equations. The AM-GM inequality is used to show that $pqrs=81$ is the only possible solution, and therefore $p=q=r=s=3$ is the final answer.
  • #1
anemone
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MHB
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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 $p=q=r=s=3$?
 
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  • #2
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.
 
  • #3
anemone said:
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 $p=q=r=s=3$.
 
  • #4
Ackbach said:
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.

Opalg said:
...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!
 
  • #5


Yes, you are correct. Since $pqrs$ must satisfy both inequalities, the only possible value is $81$. This means that $p=q=r=s=3$ is the only solution to the given system of equations.
 

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