I came across the following inequality in "Classical and New Inequalities in analysis" by Mitrinivic and Pecaric. The inequality is stated as follows.
Suppose that $F$ is an increasing function on $[a,b]$ and $F$,$G$,$H$ are integrable functions on $[a,b]$ such that for all $x$ in $[a,b]$ we have
$$ \frac{\int_a^x H(t)dt}{\int_a^b H(t)dt}\geq \frac{\int_a^x G(t)dt}{\int_a^b G(t)dt} \qquad(*)$$
Then,
$$ \frac{\int_a^x H(t) F(t)dt}{\int_a^b H(t)dt}\leq \frac{\int_a^x G(t) F(t)dt}{\int_a^b G(t)dt} \qquad (**)$$
I was able to prove $(**)$ under additional assumptions (such as $H,G$ are continuous or $F$ is continuously differentiable), but I am struggling to relax these conditions. For example, when $H,G$ are assumed continuous then $(*)$ implies that $H(x)/\int_a^b H(t)dt\geq G(x)\int_a^b /G(t)dt$ for each $x\in[a,b]$ and then the inequality follows by noting that
$$(F(b)-F(x)) \left(\frac{H(x)}{\int_a^b H(t)dt}- \frac{G(x)}{\int_a^b G(t)dt}\right) \geq 0$$
When $F$ is integrable, I used a different approach with changing the order of integration of
$$\int_{x=a}^b F'(x) \left(\frac{\int_{a}^{x}H(t)dt}{\int_a^b H(t)dt}- \frac{\int_{a}^{x}G(t)dt}{\int_a^b G(t)dt}\right)dx \geq 0$$
But I am not sure how to proceed when $G,H$ are not continuous or when $F$ is not integrable. My guess is that I need to show that
$$ H(x)/\int_a^b H(t)dt\geq G(x)\int_a^b /G(t)dt \quad a.e. $$
holds even when $H,G$ are not continuous, but I am not sure how to do so, or if this is correct.
Best Answer
It is false. Counterexample: $$F\equiv 1,G\equiv \frac1{b-a}, H=G+f, f=\chi_{[a,\frac{a+b}2)}-\chi_{[\frac{a+b}2,b]}.$$ In this case, $F$ is an increasing function on $[a,b]$ and $F$,$G$,$H$ are integrable functions on $[a,b]$, and $\int_a^b G=\int_a^b H=1$ so $$\frac{\int_a^x H(t)dt}{\int_a^b H(t)dt}- \frac{\int_a^x G(t)dt}{\int_a^b G(t)dt}=\int_a^xf(t)\,dt\geq 0\ \text{for all }x\in[a,b],$$ but $$ \frac{\int_a^{\frac{a+b}2} H(t) F(t)dt}{\int_a^b H(t)dt}- \frac{\int_a^{\frac{a+b}2} G(t) F(t)dt}{\int_a^b G(t)dt}=\int_a^{\frac{a+b}2}f(t)\,dt=\frac{b-a}2>0. $$