Comparison between two quotients of integrals involving general functions.

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?

• 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.
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.