"In Between" meaning that:

if m/n < √2 then √2 < (m+2n) / (m + n)

if m/n > √2 then √2 > (m+2n) / (m + n)

I get that by random calculations. For example:

3/2 = 1.5, (3+2×2)/(3+2) = 1.4

Wow!

My phraseology may not be the best here, but you get the idea. The proof is easy indeed, I could do it in my head (which is my excuse for not having it as polished as possible here).

It is obvious that for any given natural numbers m and n, either

or

is true, since √2 is irrational and therefore cannot be equal to a rational number, and thus the result follows from trichotomy.

Now consider the inequality

where the ? denotes the unknown direction of the inequality. Since m and n are natural numbers, their sum is nonzero and positive, and thus we may multiply each side my m + n without reversing the inequality:

Once again, n is a natural number and thus is nonzero and positive, so we may divide each side by n and maintain the direction of the inequality:

Now subtract m/n and √2 from each side to obtain

which gives (after dividing through by 1 - √2, flipping the inequality, rationalizing, then flipping again so that we now have the original direction (these steps are left to the reader as an exercise))

which gives the direction of the inequality as the opposite of the "initial condition" m/n ¿ √2, where ¿ denotes the original (and opposite) inequality direction, so that if m/n < √2 then (m + 2n)/(m + n) > √2 or if m/n > √2 then (m + 2n)/(m + n) < √2.

Thus, either

or

is true for natural numbers m and n.

I tried following through Zhylliolom's method with k instead of 2, and I'm pretty sure you're right.

For values such as 4, 9, 16, etc., Zhylliolom's first step (saying either m/n > √k or m/n < √k) breaks down.
However, if such a value is chosen, and m and n are chosen such that m/n = √k, then we can still use Zhylliolom's method to show that (m+kn)/(m+k) will also equal √k, so your corollary still holds.

Very interesting problem though, Ricky. What was that book called?

Very interesting problem though, Ricky. What was that book called?

The book is called "Analytic Inequalities", by Nicholas D. Kazarinoff.  It's aim is to fill a gap left between algebraic (high school) math and analysis, specifically that of inequalities.  Analysis is just full of them, and many students don't have a lot of experience with them.  I was one of those.  I did just find in my introduction to analysis course, however did struggle a lot with inequalities.

I will use this thread to post some of the interesting problems I find in it (or ones I have difficulty with).

Problem 1.2

Show that:

There is a very crucial hint given, but I will wait a day or two to post it.  Note that no computer nor calculator may be used.

Here is your hint:

First prove that:

As with most proofs on the natural numbers, the recommended way to proceed is by induction.

We know that:

So:

and so:

Thus:

And so:

which means that:

And so:

Whew!  The other side is exactly the same, just start off with the fact that: