# Upsetting Integral inequality

I have two smooth scalar non-negative functions f,g in $\mathbb{R}^+$ such that $$f(x) \leq g(x), ~\forall x\in \mathbb{R}^+$$ which are integrable with finite integrals in $\mathbb{R}^+$. I would like to prove that $$\frac{\int_{0}^{\infty} \frac{x}{1+x} f(x) dx}{\int_{0}^{\infty} \frac{x}{1+x} g(x) dx}\geq \frac{\int_{0}^{\infty} f(x) dx}{\int_{0}^{\infty} g(x) dx}$$ Any suggestion?

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Try the discrete version (use sums instead of integrals), to see where it leads you. –  Alecos Papadopoulos Nov 6 '13 at 10:56
I tried this with $f(x)=x,g(x)=x^2$ on the interval $[1,t)$, and the inequality was in the opposite direction from what you now have. In other words, the smaller function goes with the smaller side of the inequality between integral ratios. [naturally I truncated $f,g$ to $[0,t]$ in calculating the integrals.] Are you sure you have the direction correct? –  coffeemath Nov 6 '13 at 20:25

I don't think the direction of the inequality is something that follows from the assumption that $f(x)\le g(x),$ so I will use $\Delta$ for either of the relations $\le$ or $\ge$ supposed to hold between the ratios. Since the integrals are all positive, the relation may be written as $$\frac{\int_{0}^{\infty} \frac{x}{1+x} f(x) dx}{\int_{0}^{\infty} f(x) dx}\Delta \frac{\int_{0}^{\infty} \frac{x}{1+x} g(x) dx}{\int_{0}^{\infty} g(x) dx}.$$ Using that $x/(1+x)=1-1/(1+x)$ and algebra, this is equivalent to $$\frac{\int_{0}^{\infty} \frac{1}{1+x} g(x) dx}{\int_{0}^{\infty} g(x) dx}\Delta \frac{\int_{0}^{\infty} \frac{1}{1+x} f(x) dx}{\int_{0}^{\infty} f(x) dx}. \tag{1}$$ This is only a slight simplification, but I found it useful. Note that the larger function $g$ now appears on the left of the (unspecified) relation.

Since integral examples are hard to calculate, I decided to look at the discrete case, in fact the case wherein there are only two input numbers $x_1,x_2$ at which $f,g$ are positive, and one replaces integrals by sums over the two terms.

It turns out there is no relation between the discrete versions of these ratios of sums. So I suspect the integral version is not correct either, since one may approximate discrete sums by integrals using smooth functions sharply spiked near the inputs of the discrete functions.

The discrete version of $(1)$ is then, if $f(x_i)=a_i$ and $g(x_i)=b_i$ for $i=1,2$, as follows:

$$\frac{b_1/(1+x_1)+b_2/(1+x_2)}{b_1+b_2} \Delta \frac{a_1/(1+x_1)+a_2/(1+x_2)}{a_1+a_2}.$$ Multiplying through by the positive quantity $(1+x_1)(1+x_2)$ and subtracting $1$ from the sides brings this to $$\frac{b_1x_2+b_2x_1}{b_1+b_2} \Delta \frac{a_1x_2+a_2x_1}{a_1+a_2}. \tag{2}$$ Recall that all we have assumed is that $a_1 \le b_1$ and $a_2 \le b_2,$ this being the discrete version of $f(x)\le g(x).$ But the expressions on either side of $(2)$ are only weighted sums of the real numbers $x_1,x_2$, and the numbers $a_i,b_i$ may be so chosen as to cause either side of $(2)$ to exceed the other, even while retaining the requirements $a_i \le b_i.$

We conclude as claimed that there can be no relation, either $\le$ or $\ge$, which can serve to fill in the symbol $\Delta$ used above between the (discrete versions of) the two sides of any proposed inequality.

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