Thread: Geometric proof that the geometric mean is less than or equal to the arithmetic mean.

hello everyone,

I have been having a bit of trouble answering the question in the image, I am trying to prepare for an exam and have no one to help.

Thank you, your help is appreciated.

Originally Posted by kristina12

hello everyone,

I have been having a bit of trouble answering the question in the image, I am trying to prepare for an exam and have no one to help.

Thank you, your help is appreciated.

hello kristina!
welcome to the forum.
observe that triangle PAM is similar to triangle PMB. now can you finish?

I hate to sound stupid, but how would I go about it from here, its the first time I've seen a question like this. Im not sure what step to take?

Originally Posted by kristina12

I hate to sound stupid, but how would I go about it from here, its the first time I've seen a question like this. Im not sure what step to take?

are you familiar with "basic proportionality theorem"??
since triangles APM and MPB similar hence:
.
now do it.

PM=MB x AM
________
2

Does this prove the arithmetic mean?

So then square root AM x MB= PM

Does this prove the Geometric mean?

PM=MB x AM
________
2

Does this prove the arithmetic mean?

So then square root AM x MB= PM

Does this prove the Geometric mean?

Originally Posted by kristina12

PM=MB x AM
________
2

Does this prove the arithmetic mean?

So then square root AM x MB= PM

Does this prove the Geometric mean?

to prove AM>GM consider the centre of the circle(call it C) and join it with the "top most" point of the circle in the figure. call the top most point as K. then you can easily see from the figure that CK>PM. CK=radius of the circle. can you prove that the radius of the circle is (x+y)/2 (which is the arithmetic mean of x and y)??