$\int_{a}^{\frac{a+b}{2}}\frac{g(x)}{(1-x)^2}dx\geq\int_{\frac{a+b}{2}}^{b}\frac{g(x)}{(1-x)^2}dx$?

Question

Suppose $g>0$, $g$ decreasing on $[a,b]$ where $0<a<b<1$. Is it true or false that $$\displaystyle\int_{a}^{\frac{a+b}{2}}\frac{g(x)}{(1-x)^2}dx\geq\int_{\frac{a+b}{2}}^{b}\frac{g(x)}{(1-x)^2}dx ?$$

Answer 1

We use the following stronger Lemma which is easy to prove:

Lemma:I Let $f$ be an strictly increasing function. Then $$\int_{a}^{a+L}f(x)\, dx<\int_{b}^{b+L}f(x)\, dx \forall L>0\iff a<b$$

Now the main point is to see that we can apply this lemma in your case as both intervals share the same length. You just need to chose $g$ such that $\frac{g}{(1-x)^2}$ is strictly increasing: for example $g=1$ or $g=1-x$

Answer 2

The inequality does not hold. Counterexample: $g(x)=1$ and $0 \lt a=\frac{1}{4} \lt b=\frac{3}{4} \lt 1$ then:

$$\int_{1/4}^{1/2}\frac{dx}{(1-x)^2} = \frac{1}{1-x} \bigg|_{1/4}^{1/2} = \;\frac{2}{3} \quad \lt \quad 2\; = \frac{1}{1-x} \bigg|_{1/2}^{3/4} = \int_{1/2}^{3/4}\frac{dx}{(1-x)^2}$$