anemone
Gold Member
MHB
POTW Director
- 3,851
- 115
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 $p=q=r=s=3$?
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 $p=q=r=s=3$?