# The ratio of two strictly increasing functions

although it seems very simple and obvious, I have no idea how to give an analytical proof for this problem. I will be very happy if there are some smart ideas...

Given,

$f_1(a), f_2(a),..., f_n(a)$ and $g_1(a), g_2(a),..., g_n(a)$ are strictly increasing positive functions of $a$.

It is also known that

$\frac{f_1(a)}{g_1(a)}$, $\frac{f_2(a)}{g_2(a)}$,...,$\frac{f_n(a)}{g_n(a)}$ are strictly increasing functions of a.

I want to know if

$$\frac{f_1(a)+f_2(a)+...+f_n(a)}{g_1(a)+g_2(a)+...+g_n(a)}$$

is also an increasing function of $a$.

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It seems not obvious for me. – 89085731 Apr 25 '12 at 11:39
May be you are right. I thought that $f_i$ are greater and $g_i$ are smaller. As a result,the whole nominator of the last expression would be also greater than the expression in the denominator.. I dont know.. – Seyhmus Güngören Apr 25 '12 at 11:56
Also on MO. If you post on different fora, please mention that you did so in order to avoid duplication of efforts. – t.b. Apr 25 '12 at 12:07
I asked them to close it there. Thanks alot. They suggested me to come to this web-site. – Seyhmus Güngören Apr 25 '12 at 12:09

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Thank you very much. It is false. It is neither incereasing nor decreasing and no such generalizations can be made. – Seyhmus Güngören Apr 25 '12 at 12:10

No. A counterexample is

• $f_1(a) = 2a+a^2$
• $f_2(a) = a^3$
• $g_1(a) = a$
• $g_2(a) = a^2$

Where $$\frac{2a+a^2+a^3}{a+a^2} = \frac{2+a+a^2}{1+a}$$ decreases from $a=0.1$ to $a=0.2$.

Intuitively, the problem is that $\frac{f_1+f_2}{g_1+g_2}=\frac{f_1}{g_1}\frac{g_1}{g_1+g_2} + \frac{f_2}{g_2}\frac{g_2}{g_1+g_2}$ is a weighted average of $\frac{f_1}{g_2}$ and $\frac {f_2}{g_2}$, where we don't have any applicable hypotheses about how the weights evolve. If $g_2$ grows faster than $g_1$, then the weight shifts from $\frac{f_1}{g_1}$ for $\frac{f_2}{g_2}$, which can make the average drop of the latter is smaller and changes slowly compared to the relative change between $g_1$ and $g_2$.

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Hint:

For $a,b,c,d > 0$ we have something I call naive fraction addition rule:

$$\frac{a}{b} \leq \frac{c}{d} \Rightarrow \frac{a}{b} \leq \frac{a+c}{b+d} \leq \frac{c}{d}$$

Edit:

Ok, I made a mistake in my calculations, checking it for the second time I created a counter-example, so your conjecture is false:

Let $f_1(x) = 1+x$, $f_2(x) = x^3$, $g_1(x) = 2+x$, $g_2(x) = 100x^2$, then $$h(x) = \frac{f_1(x)+f_2(x)}{g_1(x)+g_2(x)} = \frac{1+x+x^3}{2+x+100x^2}$$ is not increasing, in fact $h(0) = \frac{1}{2}$ and $h(1) = \frac{3}{103} < \frac{1}{2}$. I know that those functions are not strictly positive and increasing on whole $\mathbb{R}$, but I guess it is easy enough to see how to fix those minor issues (and writing it explicitly here would make the example unreadable).

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this might be very useful. I think there should be also a more general version of this such that i will have $a_1$, $a_2$,...,$a_n$ instead of $a,b,c,d$. – Seyhmus Güngören Apr 25 '12 at 11:51
A more interesting question is, given the same conditions in the OP: Is (f_1 + ... +f_n)/(g_1 + ... +g_n) EVENTUALLY increasing? I have tried a few ideas with tanh(x) and 1/x^n's, but can't find anything. – Adam Rubinson Apr 25 '12 at 12:56
Yes Adam, It seems much more interesting. I tried some polynomials and all were eventually increasing. Did you find any counterexample where it doesnt increase eventually? – Seyhmus Güngören Apr 25 '12 at 17:00
@SeyhmusGüngören actually, it is enough to take example by scineram and rescale the domain. – dtldarek Apr 25 '12 at 17:37