Is there a concise way to prove that $\frac{ab}{a+c} \in [0, 1]$ for all $a > 0$, $b \in [0, 1]$, and $c \in [0, 1]$?
Proof that $ab/(a+c)$ is bounded for bounded $b$ and $c$
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proof-writing
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5Note that $0\leq ab\leq a\leq a+c$. Then you can get the conclusion. – 2012-05-11
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0@molan Perfect, thanks! Make it an answer and it gets my vote. – 2012-05-11
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0No thanks. But I think it don't deserve that. – 2012-05-11
1 Answers
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$$0\leq \frac{ab}{a+c}=\frac{b}{1+\frac{c}{a}}\leq b\leq 1$$
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0You could add $0 \le$ on the left hand side and $\le 1$ on the right – 2012-05-11
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0You also should assume that $a\not=0$. – 2012-05-12
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0It is given that $a>0$ – 2012-05-12