Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

## #1 2005-11-26 02:35:21

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 2005-11-26 04:02:52

ryos
Power Member

Offline

### 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.

## #3 2005-11-26 08:49:29

kylekatarn
Power Member

Offline

### Re: Hard Question! Need help!

I'm proving this now.

Proof by student hipnotization:

You are now under my control......

Belive me.... The statement TRUE!

Ohhhhmmmmmmm.....

3...2...1...

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

*Any questions?*

## #4 2005-11-26 09:33:09

MathsIsFun

Offline

### 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