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I am thinking if the following condition is in general true: $\frac{n}{m} \leq \frac{\sum_{i = 1} ^ {n} a_i}{\sum_{i = 1} ^ {m} b_i}$, when $n \leq m$ and $a_i \geq 0$, $b_i \geq 0$ but i cannot find a proof.

Can you please help me with that?

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Why on earth would that be true? Replacing $a_i$ by $\epsilon a_i$ for $\epsilon>0$, you can make the right hand side as close to zero as you might wish. – Harald Hanche-Olsen Sep 14 '12 at 15:08
Or what about $a_i=1, b_i = 2$? You might be able to salvage your conjecture by imposing some additional conditions on the $a$s and $b$s, but as it stands, it's just wrong. – Rick Decker Sep 14 '12 at 15:21

The condition you mentioned is false. In particular, since you let $a_i = 0$ if we want, let $a_i = 0$ for all $i$, and let $b_i = 1$ or any non-zero number. Then the right side is $0$.

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Yes! ... you are right. Thank you! – Bogdan Sep 17 '12 at 6:37

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