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Suppose I have two positive numbers $x, y$.

Obviously $$q=\dfrac{x}{x+y} \in (0, 1)$$ since $x + y > x$, implying $\dfrac{x+y}{x} > 1$ and thus $\dfrac{x}{x+y} < 1$. This is bounded below by $0$ is because a quotient of positive numbers is positive (or maybe there's some other reason, but this is irrelevant for the crux of my question).

Consider two cases: $x > y$ and $x < y$. What happens to $q$?

Apparently $q \in (0, 1/2)$ in one case, and $q \in (1/2, 1)$ in another, but it isn't clear to me how this would be shown. Particularly, how is this bound of $1/2$ derived?

I suppose we could take two positive numbers, say $3 > 2$ and set $y = 3$ and $x = 2$. Then $q = \dfrac{2}{5} \in (0, 1/2)$ - so this suggests that when $y > x$, then $q \in (0, 1/2)$ (most likely). If we take $x = 3$ and $y = 2$, then $q = \dfrac{3}{5} = 0.6 \in (1/2, 1)$. So this suggests $$q \in (0, 1/2)\text{ when } y > x$$ and $$q \in (1/2, 1) \text{ when } x > y$$ but I can't think of any way of actually showing this.

2 Answers 2

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Hint :-

Case 1. When $y>x\implies \frac{x}{x+y}<\frac{x}{x+x}=\frac{1}{2}$.

Similarly the other case follows.

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    So, it looks like I should go to bed. I'll mark your answer as best in the morning. Thank you! – 2017-01-03
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    @Clarinetist glad to help !! – 2017-01-03
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    Sorry, what do you do for the other case? We have x/(x+y) > y/(x+y), then – 2017-01-03
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    Sorry, figured it out. Thanks! – 2017-01-03
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Hint

$$\frac{x}{x+y} +\frac{y}{x+y} =1 \\ \frac{x}{x+y} > \frac{y}{x+y} \Leftrightarrow x >y$$