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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.

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Since $x>y$ and both numbers are positive, $$ \frac{x}{x+y}>\frac{x}{x+x}=\frac{x}{2x}=\frac{1}{2}. $$

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    Justification for $\frac{x}{x + y} > \frac{x}{x + x}$: All numbers are positive here. And if $x + y < x + x$, then $\frac{1}{x + y} > \frac{1}{x + x}$. Now just multiply both sides by the positive number $x$. – 2017-01-03
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    yes, I was mistaken – 2017-01-03
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    Oops, ignore my comment. Sorry. – 2017-01-03
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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, $$\frac{x}{x+y} = 1 - \frac{y}{x+y} > 1 - \frac{y}{2y} = \frac{1}{2}.$$

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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?