Show $\frac{x}{x+y} \in (1/2, 1)$ when $x > y$ Follow up to Show $\dfrac{x}{x+y} \in (0, 1/2)$ or $(1/2, 1)$ depending on whether $x < y$ or $y > x$.
It was late at night when I wrote the question above up, so I thought I understood how to do the second part of the proof, but it turns out that I don't.

Let $x, y > 0$ be such that $x > y$. Show $\dfrac{x}{x+y} \in (1/2, 1)$.

I'm not concerned with the upper bound of $1$ so much as the lower bound of $1/2$.
If $x > y$, then
$$\dfrac{x}{x+y} > \dfrac{y}{x+y}$$
but I can't show that this is greater than $1/2$ by "changing" the $x+y$ denominator to $y+y$, since
$$\dfrac{1}{x+y} > \dfrac{1}{2y}\Longleftrightarrow2y > x+y\Longleftrightarrow y > x$$
whcih is obviously not true.
 A: Since $x>y$ and both numbers are positive, 
$$
\frac{x}{x+y}>\frac{x}{x+x}=\frac{x}{2x}=\frac{1}{2}.
$$
A: Let $x,y \in \mathbb{R}$ with $x,y > 0$. $$\frac{x}{x+y} = 1 - \frac{y}{x+y} < 1$$ and if $x > y$ then, $$\frac{x}{x+y} = 1 - \frac{y}{x+y} > 1 - \frac{y}{2y} = \frac{1}{2}.$$
A: Well, if $x > y$, then $\frac{1}{2} x > \frac{1}{2} y$, right?
Then add $\frac{1}{2}x$ to both sides of the inequality, giving:
$$\frac{1}{2}x + \frac{1}{2}x > \frac{1}{2}x + \frac{1}{2} y $$
which becomes
$$ x > \frac{1}{2}( x + y) $$
and dividing by $(x + y)$, which is positive and so doesn't change the direction of the inequality, gives
$$\frac{x}{x + y} > \frac{1}{2}. $$
Remark.  I came up with this proof in reverse.  I wrote down what we wanted first, which was $\frac{x}{x + y} > \frac{1}{2}$, then manipulated this with "reversible" operations in hopes that I would get $x > y$, which we assumed.  When I was able to get that, then I just wrote up the proof backwards starting with $x > y$ and ending with what we wanted.  But to derive the proof, I actually started in reverse.  Does that make sense?
