Question

how to prove (a_n+b_n)/2 > sqrt(a_n*b_n) using induction?

Answer 1

http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#The_inequality

Answer 2

I'm not the master of induction, but I can try...
are there any restrictions on \(a_n\) and \(b_n\) ?

Answer 3

@eliassaab that's a nice generalization, but not a proof by induction

Answer 4

a_n+1=1/2(a_n +b_n)
b_n+1=sqrt(a_n x b_n)

Answer 5

i understood the relation between those two. but i cant solve the proof using induction

Answer 6

Read this from wikipeia
In two dimensions, 2x1 + 2x2 is the perimeter of a rectangle with sides of length x1 and x2. Similarly, 4√(x1x2) is the perimeter of a square with the same area. Thus for n = 2 the AM-GM inequality states that a square has the smallest perimeter among all rectangles with equal area,

Answer 7

again, not inductive...

Answer 8

@eliassaab , tq for that. i understand about that am-gm theory, but i need the induction part of proving

Answer 9

For any two positive numbers a and b
\[
\sqrt{a b} \le \frac { a+b} 2
\]
There is no need for induction here.

Answer 10

ok i see that this is probably my fault, because i mentioned induction earlier. forget that.
it is always the case that \[\frac{a+b}{2}\geq \sqrt{ab}\] the subscript n is irrelevant

Answer 11

but that's what they asked for...

Answer 12

is this a follow up Q ?

Answer 13

no i think that was because i stupidly mentioned it in the answer to a previous problem, which his what i get for writing before thinking clearly

Answer 14

ah, ic

Answer 15

forget i mentioned it, it was stupid on my part

Answer 16

nobody is stupid. we learn n learn....

Answer 17

@TuringTest original question was here
http://openstudy.com/study#/updates/4f8eb037e4b000310face9e5

Answer 18

means, we can just explain like that, and no need for induction part?