# AM-GM inequality proof helpWatch

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Thread starter 1 year ago
#1

Hi, I was trying to rediscover this ingenious proof by Cauchy, but couldn't think of any clever substitution in part (b).
Things I have tried:

Spoiler:
Show

a=4x/3, b=4y/3, c+d=4z/3 gives cd=9/16 z(xyz)^1/3. Ugly!! Showing existence of positive solutions for c and d drove me nuts
I am pretty sure there are nicer solution.
Spoiler:
Show

a=(x)^4/3, b=(y)^4/3 and (cd)=z^4/3

Could anyone give me a hint or two for part b, if it is possible to give a hint without telling the answer?

I think I will be able to do part (c) once I get the part (b) right but I could be wrong.

Thank you!!
0
1 year ago
#2
So, I tried to make a small hint for this, and realised that I'd actually made a massive assumption up front (that works, but counts as a fairly big hint in it's own right). So spoilering the whole thing now:

Spoiler:
Show
The big hint here is that you can leave a,b,c unchanged, and just choose a suitable d.

So, you want to go from knowing

to

by choosing a suitable value for d.

It's hard to hint without giving away a lot but consider this:

You want to choose d as a function of a, b, c.

Because of symmetry, a symmetric function of a, b, c seems a good guess.

You know the AM-GM expression gives equality when a = b = c.

Bigger hint follows:

Spoiler:
Show

Consider working backwards from the 2nd expression to the first. How might you change that "^3" into a "^4"?
2
1 year ago
#3
FWIW, I never got this without a huge hint. And I even considered the correct choice, but I just "looked briefly" and thought "that doesn't actually work" when in fact it does.
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Thread starter 1 year ago
#4
(Original post by DFranklin)
So, I tried to make a small hint for this, and realised that I'd actually made a massive assumption up front (that works, but counts as a fairly big hint in it's own right). So spoilering the whole thing now:
Thanks for the hint. Haven't looked at the bigger hint (second spoiler) yet but knowing that d is symmetric function of a,b,c made made me think it could well be a function of a+b+c and Bingo!!! X=a+b+c and solving X/3 = X+d/4 gave me d=(a+b+c)/3 and part (b) and (c) become trivial.
Really appreciate it!!
0
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