proof for arithmetic mean>=geometric mean? (1 Viewer)

[u-borat]

there is a proof in the 1998 hsc q8, but it relies on previous parts.

that begs the question, would the BOS ask for this proof without any hints or is it too difficult?

 

[bazza159]

its not too hard

Q: Prove [a - b]/2 >= root ab, a>0, b >0

Now (root a - root b)^2 >= 0

a - 2root ab + b >= 0

a + b >= 2root ab

(a+b)/2>= root ab

hence arithmetic mean >= geometric mean

 

[shaon0]

there is a proof in the 1998 hsc q8, but it relies on previous parts.

that begs the question, would the BOS ask for this proof without any hints or is it too difficult?

No its not that hard (as shown in the above post). Our teacher did it in class in a minimal amount of steps. So you'd probably be able to do it.

 

[u-borat]

does that proof cover n values?

 

[Affinity]

you can do the n value case by induction.

There are plenty of other ways to do it though..
I think the paper you refered to essentially used holders or jensens inequality

 

[AMorris]

you can do the n value case by induction.

The proofs that are solely by induction (i.e. no calculus) are quite tricky and the two common ones which are listed on the wikipedia article have a few not so straightforward steps. It is easier to prove it using calculus and fiddling around with maxima and minima.

Fortunately, they will never ask you to prove more than the n=3 case without giving you a complete walk through of the proof so there is absolutely no need to memorise the proof of the general case.

 

[u-borat]

ok thanks that is what i was hoping to hear.