I've been thinking about inequalities of the kind:

$$\frac{\sum_i a_i}{\sum_i b_i}\leq \sum_i \frac{a_i}{b_i} \qquad a_i,b_i>0$$

For some cases, it is easy to see why it works, for example:

$$\frac{a_1+a_2}{b_1+b_2}\leq \frac{a_1}{b_1}+\frac{a_2}{b_2}\qquad a_i,b_i>0$$

We write

$$0\leq a_1b_2^2+a_2b_1^2$$

And then:

$$a_1b_1b_2+a_2b_1b_2 \leq a_1b_1b_2+a_2b_1b_2+ a_1b_2^2+a_2b_1^2$$


$$b_1b_2(a_1+a_2)\leq (a_1b_2+a_2b_1)(b_1+b_2)$$

Which yields the desired inequality:

$$\frac{a_1+a_2}{b_1+b_2}\leq \frac{a_1}{b_1}+\frac{a_2}{b_2}$$

I am curious about the following:

  • How can we find a good upper bound for $\sum_i \frac{a_i}{b_i}$ in terms of $\frac{\sum_i a_i}{\sum_i b_i}$? In my hypothetical application for this, obtaining the number $\frac{\sum_i a_i}{\sum_i b_i}$ is very easy but obtaining $\sum_i \frac{a_i}{b_i}$ would be very messy. Having a good upper bound in terms of $\frac{\sum_i a_i}{\sum_i b_i}$ would be desirable. I know next to nothing about inequalities. I guessed about squares, cubes of $\frac{\sum_i a_i}{\sum_i b_i}$ but this would perhaps be too large. I guess I don't know how to find better upper bounds in terms of $\frac{\sum_i a_i}{\sum_i b_i}$.
  • $\begingroup$ Titu's Lemma ${}$ $\endgroup$
    – TheSimpliFire
    Nov 14, 2020 at 13:51
  • $\begingroup$ @TheSimpliFire No, it is not. $\endgroup$ Nov 14, 2020 at 16:39
  • $\begingroup$ @Billy Rubina: What are you expecting? It's obvious after expanding $\endgroup$ Nov 14, 2020 at 16:42
  • $\begingroup$ Are you looking for an upper bound or lower bound for $\sum \frac{a_i}{b_i}$? In the $a_1, a_2$ example you gave, its a lower bound that's shown. A lower bound is easy using CS inequality. $\endgroup$
    – Macavity
    Nov 14, 2020 at 17:15
  • $\begingroup$ An upper bound isnt likely, just consider any one $b_i$, say $b_1 \to 0^+$. $\endgroup$
    – Macavity
    Nov 14, 2020 at 17:23

1 Answer 1


If you want an upper bound that only depends on $\frac{\sum a_i}{\sum b_i}$ then no.

The quantity $\frac{\sum a_i}{\sum b_i}$ is called the mediant, or freshman sum, of the fractions $\frac{a_i}{b_i}$. It is a weighted average of the fractions so it's always between $\min \frac{a_i}{b_i}$ and $\max \frac{a_i}{b_i}$ and can be arbitrarily close to either. And this is exactly how Simpson's Paradox works.


Now if $n=2$, WLOG assume $\frac{a_1}{b_1} < \frac{a_2}{b_2}$. The mediant can be arbitrarily close to $\frac{a_1}{b_1}$ while the ratio of $\frac{a_2}{b_2}$ to $\frac{a_1}{b_1}$ can also be arbitrarily large. Therefore $\frac{\sum \frac{a_i}{b_i}}{\frac{\sum a_i}{\sum b_i}}$ can be arbitrarily large.

For example, if $N$ is very large, then $$ \frac{\frac 1N + \frac{N}{1}}{\frac{1+N}{N+1}} = N+\frac 1N > N. $$

  • $\begingroup$ This is nice. My plan here was trying to find an upper bound for that quantity I judged harder to compute, do you think there is any other way to find a good upper bound for it? $\endgroup$
    – Red Banana
    Dec 9, 2020 at 6:17
  • 1
    $\begingroup$ Can't think of anything without additional information about the $a$'s and $b$'s. $\endgroup$
    – Neat Math
    Dec 9, 2020 at 14:30
  • $\begingroup$ What kind of information would yield nice results? Can you think on something? $\endgroup$
    – Red Banana
    Dec 10, 2020 at 2:53
  • 1
    $\begingroup$ Say if $m\le b_i \le M$ then $\sum a_i/b_i \le \frac Mm \frac{\sum a_i}{\sum b_i}$ $\endgroup$
    – Neat Math
    Dec 10, 2020 at 15:00
  • 1
    $\begingroup$ Thanks for your kind words. Back in high school and college I was lucky enough to have a few genius friends. I learned more from them than from the teachers/professors. Haven't studied any particular inequality books. I think you can just look around and find good peers to learn from. There are many experts in inequality here at Math SE as well. Good luck. $\endgroup$
    – Neat Math
    Dec 11, 2020 at 1:31

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