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

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The mathematical problem involves determining the values of positive real numbers p, q, r, and s that satisfy the equations p + q + r + s = 12 and pqrs = 27 + pq + pr + ps + qr + qs + rs. Using the AM-GM inequality, it is established that pqrs must equal 81, leading to the conclusion that p, q, r, and s must all be equal to 3. This conclusion arises because equality in the AM-GM inequality occurs only when all variables are equal. Although the solution is confirmed, the discussion raises the question of whether other solutions exist, suggesting further exploration with different initial values. Ultimately, the consensus is that p = q = r = s = 3 is the only solution that meets the given conditions.
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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:

[TABLE="class: grid, width: 700"]
[TR]
[TD]1.[/TD]
[TD]$\dfrac{p+q+r+s}{4} \ge \sqrt[4]{pqrs}$ which then gives $(\dfrac{12}{4})^4 \ge pqrs$ or $pqrs \le 81$[/TD]
[/TR]
[TR]
[TD]2.[/TD]
[TD]$\dfrac{pq+pr+ps+qr+qs+rs}{6} \ge \sqrt[6]{(pqrs)^3}$

which then gives $(\dfrac{pqrs-27}{6})^2 \ge pqrs$

$(pqrs-81)(pqrs-81) \ge 0$

$pqrs \le 9$ or $pqrs \ge 81$[/TD]
[/TR]
[/TABLE]

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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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.
 
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:

[TABLE="class: grid, width: 700"]
[TR]
[TD]1.[/TD]
[TD]$\dfrac{p+q+r+s}{4} \ge \sqrt[4]{pqrs}$ which then gives $(\dfrac{12}{4})^4 \ge pqrs$ or $pqrs \le 81$[/TD]
[/TR]
[TR]
[TD]2.[/TD]
[TD]$\dfrac{pq+pr+ps+qr+qs+rs}{6} \ge \sqrt[6]{(pqrs)^3}$

which then gives $(\dfrac{pqrs-27}{6})^2 \ge pqrs$

$(pqrs-81)(pqrs-81) \ge 0$

$pqrs \le 9$ or $pqrs \ge 81$[/TD]
[/TR]
[/TABLE]

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 $p=q=r=s=3$.
 
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.:o

I understand it all now! Thanks guys!
 
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