Prove "1/4a+1)+(1/4b+1)+(1/4c+1) ≥ 1" With Fun Proof

[Bitter]

First of all, This problem was proposed by Sefket Arslanagic, University of Sarevo.

Let a, b, c be positive real numbers such that

(1/a+1)+(1/b+1)+(1/c+1)=2

Prove that

(1/4a+1)+(1/4b+1)+(1/4c+1) greater than or equal to 1

 

[ceptimus]

I get:

a = (1 - bc) / (2bc + b + c)

 

[Bitter]

i'm not sure how you got your results or how it proves the statement. Can you explain this to me?

 

[ceptimus]

I got it by a bit of algebra. I don't claim it helps in proving the statement, but (if it's right, as I think it is) then it allows anyone to guess any two values for b and c, and find the corresponding value of a that satisfies the first equation.

It at least allows you to test the second formula for a few different values, to satisfy yourself that it's probably right before embarking on a proof.

 

[tehno]

Forward this to General Math subforum and I will prove it there.

 

[Gib Z]

Heres my *PROOF*

Any numbers that satisfy a b and c for the first part, have to do it for the second part or else the questions an idiot. Try 0, 1, 1. Works for both parts, the questions not stupid, the proof is complete :P

 

[mdelisio]

Good one!

I used the method of Lagrange multipliers to find the extreme of:

(1/(4a+1))+(1/(4b+1))+(1/(4c+1)) = f(a,b,c)

Subject to the constraint that

(1/(a+1))+(1/(b+1))+(1/(c+1)) = 2

This gives a = b = c = 1/2, and f(1/2,1/2,1/2) = 1

A quick check of some other points (like 0, 1, 1) shows that this is a minimum point.

 

[Gib Z]

That would work as well...