BMO2 1996 Questio 4 - Algebra Problem

Question

Let $a,b,c$ and $d$ be positive real numbers such that

$a + b + c + d = 12$

and

$abcd=27+ab+ac+ad+bc+bd+cd$.

Find all possible values of $a,b,c,d$ satisfying these equations.

I found this problem on someone's blog, where they had also given a proof using AM-GM, but there was one part of it I couldn't follow and was wondering if anyone could help me. Feel free to come up with your own proof, but the proof in question is about halfway down this page: https://mblog1024.wordpress.com/2011/11/30/bmo2/

I couldn't understand why $6\sqrt{abcd}\geq54$

Thanks in advance for any help.

Which part you are having problem understanding??! The proof is clear!! — Ben, Jul 10 '15 at 18:07

score 2 · Accepted Answer · answered Jul 10 '15 at 18:09

2

I couldn't understand why $6\sqrt{abcd}\geq54$

Note that $$abcd-27\ge 6\sqrt[6]{a^3b^3c^3d^3}=6\sqrt{abcd}.$$ Setting $t=\sqrt{abcd}$ gives you $$t^2-27\ge 6t\iff (t-9)(t+3)\ge 0\iff t\ge 9.$$

answered Jul 10 '15 at 18:09

mathlove

139,939

1

I see now. Thanks to you and everybody else who have an answer. – MadChickenMan Jul 10 '15 at 19:05

score 2 · Answer 2 · answered Jul 10 '15 at 18:15

2

Note that $abcd-27\ge 6\sqrt{abcd}\implies(\sqrt{abcd}-3)^2\ge36\implies \sqrt{abcd}\ge 9\dots$ is probably the way it should have concluded.

answered Jul 10 '15 at 18:15

Nemo

1,270

score 1 · Answer 3 · answered Jul 10 '15 at 18:11

from $a+b+c+d=12$ we get $$\frac{a+b+c+d}{4}\geq\sqrt[4]{abcbd}$$ this means $$3\geq\sqrt[4]{abcd}$$ or powerd by $4$ $$81\geq abcd$$ this is equivalent to $$54\geq abcd-27=ab+ac+ad+bc+bd+cd\geq 6\sqrt[6]{(abcd)^3}$$ from here you will the searched inequality

BMO2 1996 Questio 4 - Algebra Problem

3 Answers3