1. ## Inequality Question

Hi, I am new to the forum, and I hope I am posting this in the right place:

Question from our grade 12 exchange student:

a and b are integers such that

0 < a < b

A = sqroot (ab)

He was able to answer the first question as follows:

1) show that a < A < b

a < b

a*a < b*a

a^2 < ba

a < sqroot(ba)

therefore

a < A < b

he was unable to answer the following:

2) show that

a < 2ab/(a+b) < A

2. ## Re: Inequality Question

Originally Posted by chet
he was unable to answer the following:
2) show that
a < 2ab/(a+b) < A
By looking at $(b-a)^2>0$ we see that if $b>a$ then $b^2+a^2>2ab$.
From that we get $a+b>2\sqrt{ab}=\frac{2ab}{\sqrt{ab}}$.

Now from that $\frac{2ab}{a+b}.

Can he finish?

3. ## Re: Inequality Question

I will ask him if he can finish. Thanks!!