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