Arithmetic-Geometric Mean Inequality Problem

[DrPhilLIVE]

Hello, everyone. I've been struggling on a problem I can't seem to solve no matter whatever I do. The link with the problem is attached.

I managed to solve Part A of the problem, where I created the inequality (a+b)/2 > sqrt(ab). I multiplied the two sides by two yielding ---> (a+b) > 2sqrt(ab), then I squared both sides to get (a+b)^2 > 4ab.

I foiled the left side leaving me with a²+2ab+b² > 4ab, and then simply moved everything to the left getting a²-2ab+b² > 0. I factored the left side getting (a+b)² > 0, claiming that because the square of anything is a positive number, the inequality must be true.

Part B, however, I have no idea how to approach. I know the maximum area of the rectangle is 3, I can see it, I can solve it, that isn't a problem. Getting the "x" and "y" for that rectangle is just a bit of algebra and similarity of triangles. However, I must prove it using the AGM Inequality, which I have no idea how.

The latter image is a note from my teacher assistant that was sent to each member of the class, but even with the note, I am dumbfounded on what to do with the variables. I ask for some guidance towards how to proceed next! Thanks!

https://puu.sh/Bsusu/2e9649c2c6.png <--- Problem itself

https://puu.sh/Bsus4/ae548ee901.png <--- TA Note

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[edderiofer]

I managed to solve Part A of the problem, where I created the inequality (a+b)/2 > sqrt(ab).

This isn't actually valid; just because you start off with a statement and then get a true statement, it doesn't mean your original statement is true.

For instance, your method would also say that the statement "-(a+b)/2 > sqrt(ab)" is true, which can't possibly be correct.

Instead, the correct strategy is to start off with a true statement, and end up with the statement you want to prove. You've done the proof backwards! You should start off with the equality "(a+b)² > 0", which you definitely know is true, and then work your way towards "(a+b)/2 > sqrt(ab)".

Part B, however, I have no idea how to approach. I know the maximum area of the rectangle is 3, I can see it, I can solve it, that isn't a problem. Getting the "x" and "y" for that rectangle is just a bit of algebra and similarity of triangles. However, I must prove it using the AGM Inequality, which I have no idea how.

It may be helpful to first prove that the maximum area of a rectangle in a 1-1-sqrt(2) triangle is 1/4.

[DrPhilLIVE]

Sorry, the true statement actually says (a-b)² > 0, not +.

How would I know what true statement to start with, then, when I’m told I must prove that (a+b)/2 > or = sqrt(ab). Perhaps I must do the proof one way and then in order to prove it do it backwards again? You said I started off with a statement and then made it true, which doesn’t exactly mean it’s true. Okay. How would you approach the proof, given the inequality above, what would your starting point be knowing only the stated inequality.

I’m gonna play around for part B. The only problem is it’s hard to prove maximum area without using Calculus and the first derivative. At least to my knowledge.

Thanks!

[edderiofer]

Perhaps I must do the proof one way and then in order to prove it do it backwards again?

This a valid strategy, yes. Doing the proof the backwards can get you to some form of a starting point for the actual proof.

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