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

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:

This inequality can be proven simply by writing it as

and then

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

Hope that helps.

Always greater than or equal to 3?

I am now attempting to prove that

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:

Simplifying to:

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)