Am-gm (1 Viewer)

[Estel]

Specific q;
can it be assumed for case
n=2;
n=3;
n=4;
if it is NOT proved in the leadup.

 

[Archman]

2: proven in 2 lines, doesnt hurt to do it
3: no, can't assume it unless your told so, a short proof still takes about 10 lines
4: use the "2" case twice to prove it, not too long to do it yourself, again, not safe to assume.

 

[Estel]

ahh damn
thanks...

well one more q
can we fudge expansions...

like (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=a^3+b^3+c^3-3abc...

 

[nit]

well depending on what you want to do, LHS RHS might be all you need. But yeh, I went into the exam with that proof for the n=3 case, just to be on the safe side, and an AMGM proof that started with logx=<x-1, again just for safety, and rather useless in our exam as it was.

 

[Slidey]

Why the logx thing? I know of two ways to prove am-gm which are very short and mundane... no need for exotic logarithms.

 

[nit]

This one was pretty short - u prove logx =< x-1

Then you prove that log(np1 + np2 +...+npn)=<0, where p1 + p2 +...+pn=1
Finally you prove the AM GM with x1 + x2 +... +xn=s using the previous part. Not more than half to three-quarters of a page.

Yeh, there are way to many ways to prove AM GM...Laurie had about 6 at last count

 

[Estel]

There's a 3 line one... but you assume Jensen's

Then there's a 5 liner using differentiation that I didn't get... damn inspection laden methods!

I'll stick with the induction; AMGM for dummies

ahh well I'll read up on the log one later.
thanks

 

[Slidey]

This one was pretty short - u prove logx =< x-1

Then you prove that log(np1 + np2 +...+npn)=<0, where p1 + p2 +...+pn=1
Finally you prove the AM GM with x1 + x2 +... +xn=s using the previous part. Not more than half to three-quarters of a page.

Yeh, there are way to many ways to prove AM GM...Laurie had about 6 at last count

Half to three-quarters of a page?

(a+b)/2>=sqrt(ab), square both sides:
(a+b)^2/4-ab>=0
a^2+2ab+b^2-4ab>=0
(a-b)^2>=0 - true
.'. AM>=GM

My point is, there's no need to actually use a proof like that in an exam. Learn it for fun sure.

 

[Estel]

You've proven it for n=2...
half to three-quarters of a page is referring to the general case.

 

[turtle_2468]

Lets try listing all of them... just for fun...
1) Expansion like above, generalise to 2^n, sub back in to simplify
2) Jensen's/log
Someone help me out here...

 

[Slidey]

You've proven it for n=2...
half to three-quarters of a page is referring to the general case.

Just realised.

Here's one to add, turtle.

http://mathforum.org/library/drmath/view/51596.html

Lagrange multipliers, hoorah!

 

[Estel]

Another thing; for the induction method, how would you prove equality occurs if and only if the terms are all equal.

My book doesn't have that =/

 

[Slidey]

To prove that equality occurs iff all terms are equal, you want to prove that if not all terms are equal, then AM>GM. Then you would want to prove that if all terms are in fact equal, AM=GM.

 

[Estel]

And how would you go about doing that

 

[Archman]

4) Perturbation
5) Rearrangement

 

[Estel]

6.) Defining f(x) = AM minus GM, using minima and induction.

 

[Slidey]

7) Inspection.

 

[Archman]

anything after 6) was just ranting

 

[SeDaTeD]

Um, Tywebb, why did you post this here? You've already done so in the thoughts on hsc exam section.

 

[turtle_2468]

Rearrangement??
Good point re lagrange estel... I _am_ growing rusty..