# Math Help - Hard/beautiful inequality --help!!

1. ## 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

2. Very nice problem! Could you tell me the source of this problem? Thanks.

3. 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,

4. ## 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:

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

This inequality can be proven simply by writing it as

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

and then

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

$
\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.

5. ## Always greater than or equal to 3?

I am now attempting to prove that

$
\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:

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

Simplifying to:

$
\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?