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## #1 2005-11-25 03:35:21

math student
Guest

### Hard Question! Need help!

positive reals a,b,c give
ab + bc + ac = 3
Prove that:
a³ + b³ + c³ + 6abc ≥ 9

## #2 2005-11-25 05:02:52

ryos
Member
Registered: 2005-08-04
Posts: 394

### Re: Hard Question! Need help!

Since ab + bc + ac = 3, let ab, bc, and ac all equal 1. Now, 1³ + 1³ + 1³ + 6(1) = 9.

Is this the lower bound? My intuition says yes, because to decrease one of them, you'd have to increase at least one other, possible more than you decreased the one. However, I don't know how to prove it. Maybe someone else could step in?

El que pega primero pega dos veces.

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## #3 2005-11-25 09:49:29

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

### Re: Hard Question! Need help!

I'm proving this now.

Proof by student hipnotization:

:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P
You are now under my control......
:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|
Belive me.... The statement TRUE!
:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P
Ohhhhmmmmmmm.....
:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|:|
3...2...1...
:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P:P

Therefore, we conclude that a³ + b³ + c³ + 6abc ≥ 9
//q.e.d

*Any questions?*

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## #4 2005-11-25 10:33:09

MathsIsFun
Registered: 2005-01-21
Posts: 7,664

### Re: Hard Question! Need help!

LOL!! Gee, and I really feel like doing that with proofs ...

OK, let's just play with this, start with a=b=c=1, and then increase "a" by ε, and decrease "c" by λ to compensate

ab + bc + ac = (1+ε)1 + 1(1-λ) + (1+ε)(1-λ) = 1 + ε + 1 - λ + (1 - λ + ε - ελ)
= 3 + 2ε - 2λ - ελ = 3

∴ 2ε - 2λ - ελ = 0

So, does that "- ελ" term mean that ε has to be slightly larger than λ to make it work? And hence "a³ + b³ + c³ + 6abc" is going to increase?

I think so ...

... and I will now hypnotize you all ...

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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