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When we have the following inequality:

$$\frac{a}{b+c} \ge \frac{d}{e+c},$$

with $a,c,d \in \mathbb N_{\ge 0}$, $b, e \in \mathbb N_{\gt 0}$, $a \le b$ and $d \le e$

Then it seems to hold that

$$\frac{a}{b} \ge \frac{d}{e},$$

Is this correct? Does it also work in the other direction (iff)?

Related question : Subtracting positive numbers from denominator in an inequality.

Community
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miselico
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  • Here is a crazy suggestion: when faced to an obvious counterexample once, maybe try some similar cases before proposing a new, equally trivially wrong, conjecture. – Did Mar 24 '14 at 19:32
  • Answer posted to counter pointless hostility from the usual suspects. – zyx Mar 24 '14 at 20:17
  • @Did After comparing your counterexample with the cases which I had tried, I noticed these extra conditions. Then, I jumped too fast to conclusions. Sorry about that. – miselico Mar 24 '14 at 20:59

2 Answers2

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Like your related question, the answer to this one is no. Let $a=1, b=d=2, e=5$. (Note there is nothing special about $b=d=2$ other than they are small easy to work with numbers.)

So we have $\dfrac{a}{b} = \dfrac{1}{2} > \dfrac{2}{5} = \dfrac{d}{e}$.

If $c=3$ (though it works for any $c>1$), then $\dfrac{a}{b+c} = \dfrac{1}{5} < \dfrac{1}{4} = \dfrac{2}{8} = \dfrac{d}{e+c}$.

John Habert
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It is true that

  • $f(c) = \frac{e+c}{b+c}$ is a monotonic function of $c$ (for positive values of the other variables, and $c \geq 0$), so that

  • the sign of $(f(c) - f(0)$) will be the same for all $c > 0$,

but it is not true that function is always decreasing. That is determined by the the sign of $(e - b)$.

zyx
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  • (the question tried to infer $f(0) \geq y$ from $f(c) \geq y$, which would require $f(0) \geq f(c)$) – zyx Mar 24 '14 at 20:20