# Hard/beautiful inequality --help!!

• Jun 15th 2008, 03:59 PM
LanNguyen
Hard/beautiful inequality --help!!
Pls help to prove this inequality which I think is quite neat:

--------------
3 (a/b + b/a + b/c + c/b + a/c + c/a) + (1 + a) (1 + b) (1 + c) (c/b + c/a) (b/a + b/c) (a/b + a/c)
>=
6 abc + 6 + 9(ab + bc + ac + a + b + c) + 3(ab/c + bc/a + ac/b)
--------------

with a, b, c are positive reals

If it helps, equality occurs when a=b=c=2. Also, can any one helps to prove (2, 2, 2) is the only point at which equality occur.

Thanks
• Jul 26th 2008, 11:22 PM
mathwizard
Very nice problem! Could you tell me the source of this problem? Thanks.
• Jul 27th 2008, 03:29 PM
LanNguyen
I discovered this inequality while working on a mathematical modelling of a genetic system with negative feedback loops. I'm working in systems biology area.

I know for sure the inequality holds and can prove it computationally. I'm looking for a rigorous, mathematical proof though. Have you tried it out? Do you have any ideas on tackling it?

Cheers,
• Jul 30th 2008, 09:37 PM
Serena's Girl
Always greater than or equal to 2
I am not certain if this will be helpful, but I'm starting out with the observation that for a pair of positive reals A and B:

$\displaystyle \frac {A} {B} + \frac {B} {A} \geq 2$

This inequality can be proven simply by writing it as

$\displaystyle \frac {A^2 + B^2} {AB} \geq \frac {2AB} {AB}$

and then

$\displaystyle \frac {A^2 + B^2 - 2AB + 2AB} {AB} \geq \frac {2AB} {AB}$

$\displaystyle \frac {(A - B)^2 + 2AB} {AB} \geq \frac {2AB} {AB}$

Since (A - B)^2 will always be at least zero, the inequality holds true.

Hope that helps.
• Jul 30th 2008, 11:37 PM
Serena's Girl
Always greater than or equal to 3?
I am now attempting to prove that

$\displaystyle \frac {ab} {c} + \frac {bc} {a} + \frac {ac} {b} \geq 3$

It seems to me it is true, but I haven't found a mathematical proof for it yet. This is as far as I've gone:

$\displaystyle \frac {a^2 b^2 + b^2 c^2 + a^2 c^2} {abc} \geq \frac {3abc} {abc}$

Simplifying to:

$\displaystyle \frac {ab(ab - c) + bc(bc - a) + ac(ac - b) + 3abc} {abc} \geq \frac {3abc} {abc}$

So it's now a matter of proving that ab(ab - c) + bc(bc - a) + ac(ac - b) is at least zero. Any takers? (Lipssealed)