Q:

An elementary question about integers and fractions

Let $\mathbb{N}$ be the set of natural numbers (i.e., positive integers), and let $x, y, z \in \mathbb{N}$.  Assume further that I have $x \leq y$.
Now, suppose that I have the inequalities
$$2(y - x) + \frac{2x^2}{y + x} \leq z < 2(y - x) + \frac{2x^2 + y}{y + x}.$$
Does this force $z = 2y - x$?
Update
There is a newer question here, which more or less should have been my actual question.

A:

I noticed you have changed the problem twice now and we arrive at the third problem of a similar form. In light of this I will now attempt to answer the general case of 
$$ a \leq z < b $$
In general, the z is irrelevant. One needs only prove that the RHS is greater than the LHS and then to say that it is specifically greater than the left by a certain amount. This can be done by adding and subtracting on both sides, but a contradiction may be reached (such as z is between two consecutive integers) if multiplication and division are involved, which would imply we entered the rationals or reals.

