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Suppose we have $$\frac{a_i^1}{b_i^1+c}>\frac{a_i^2}{b_i^2+c},~\forall ~i\in\{1,2,\cdots L\}$$ and $a_i^1<a_i^2$, $b_i^1<b_i^2$ $\forall ~i\in\{1,2,\cdots L\}$. Then can we say that following inequality is true $$\frac{\sum_{i=1}^La_i^1}{\sum_{i=1}^{L}b_i^1+Lc}>\frac{\sum_{i=1}^La_i^2}{\sum_{i=1}^{L}b_i^2+Lc}?$$ Please note that $a_i,b_i$ and $c$ are non-negative values. Further please note that the superscripts do not represent powers.

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  • $\begingroup$ WLOG $c=0$, you can absorb it in the $b$'s. $\endgroup$ – Yves Daoust Jul 9 '18 at 9:13
  • $\begingroup$ @YvesDaoust yes I agree with you. But what do you think about the inequality? Is it always true? $\endgroup$ – Frank Moses Jul 9 '18 at 9:14
  • $\begingroup$ @YvesDaoust I am trying to show it for $L=2$ then will look for induction to extend the case to general $L$ case. This inequality look very simple. I am wondering whether there is some result present in the literature. $\endgroup$ – Frank Moses Jul 9 '18 at 9:25
  • $\begingroup$ @YvesDaoust Unfortunately I am unable to show it even for $L=2$ case $\endgroup$ – Frank Moses Jul 9 '18 at 9:26
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A geometrical disproof:

WLOG $c=0$, you can absorb it in the $b$'s.

Now, in a $(b,a)$ coordinate system, the individual points that fulfill the constraints are contained in an axis aligned right triangle with a vertex at the origin.

We choose a blue $p_1:=(a_1^1,b_1^1)$ and the locus of $q_1:=(a_1^2,b_1^2)$ follows. Similarly with a green $p_2:=(a_2^1,b_2^1)$.

Now take the sum of $p_1$ and $p_2$, or equivalently their arithmetic average, to get $p$. The inequation to be proven expresses that the average of $q_1$ and $q_2$ must lie below the straight line by $p$.

But it is clear that we can pick a $q_1$ and a $q_2$ such that this is invalidated.

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Finding numerical values shouldn't be a big deal.

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