A.M. >= G.M. (1 Viewer)

[~ ReNcH ~]

If you strike a question involving inequalities, are you allowed to assume that the arithmetic mean of a set of numbrs is greater than the geometric mean? i.e.

(a + b + c +.....n terms)/n >= nth rt (a*b*c*d....n terms)

 

[shafqat]

No, most probably not.
Many HSC inequality questions are themselves proofs for AM/GM in 3 terms, or similar, so assuming the result would make those questions trivial.

edit: when i finish my uni assignment tomorrow, ill post up some inequality stuff for people, like general proofs of AM/GM etc

 

[~ ReNcH ~]

So if it was necessary to use that fact, then would they most likely ask you to prove it first, in an earlier part?

 

[shafqat]

Yes although general proofs of AM/GM arent easy, and would constitute a q7/8 question on its own.

 

[withoutaface]

(a+b) / 2 >= sqrt(ab)

then (c + d) /2 >=sqrt(cd)

also (e + f )/2 >= sqrt(ef)

hence (c+d+e+f) /4 >= 4thrt(cdef)

I suppose you could do something like that if n = 2^k where k is a positive integer.

 

[~ ReNcH ~]

How do you prove AM >= GM for n=3?
I know a long way where you first prove that (x+y+z)^3 >= 27xyz and go from there. Is there any shorter method for this proof?

 

[ngai]

How do you prove AM >= GM for n=3?
I know a long way where you first prove that (x+y+z)^3 >= 27xyz and go from there. Is there any shorter method for this proof?

prove n=2
prove n=4
so (a+b+c+d)/4 >= (abcd)^(1/4) is true when d = (a+b+c)/3
expand and simplify

theres another way, which expands some crap and sorta magically gets the answer...but i cant remember it

 

[maths > english]

a generic AM >= GM proof was put in the end of session exam for math1141 (attached)