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Is there a concise way to prove that $\frac{ab}{a+c} \in [0, 1]$ for all $a > 0$, $b \in [0, 1]$, and $c \in [0, 1]$?

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    Note that $0\leq ab\leq a\leq a+c$. Then you can get the conclusion.2012-05-11
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    @molan Perfect, thanks! Make it an answer and it gets my vote.2012-05-11
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    No thanks. But I think it don't deserve that.2012-05-11

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$$0\leq \frac{ab}{a+c}=\frac{b}{1+\frac{c}{a}}\leq b\leq 1$$

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    You could add $0 \le$ on the left hand side and $\le 1$ on the right2012-05-11
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    You also should assume that $a\not=0$.2012-05-12
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    It is given that $a>0$2012-05-12