<s>[INST] $\int_{a}^{\frac{a+b}{2}}\frac{g(x)}{(1-x)^2}dx\geq\int_{\frac{a+b}{2}}^{b}\frac{g(x)}{(1-x)^2}dx$? Suppose $g>0$, $g$ decreasing on $[a,b]$ where $0<a<b<1$. Is it true or false that
$$\displaystyle\int_{a}^{\frac{a+b}{2}}\frac{g(x)}{(1-x)^2}dx\geq\int_{\frac{a+b}{2}}^{b}\frac{g(x)}{(1-x)^2}dx ?$$
 [/INST] The inequality does not hold. Counterexample: $g(x)=1$ and $0 \lt a=\frac{1}{4} \lt b=\frac{3}{4} \lt 1$ then:
$$\int_{1/4}^{1/2}\frac{dx}{(1-x)^2} = \frac{1}{1-x} \bigg|_{1/4}^{1/2} = \;\frac{2}{3} \quad \lt \quad 2\; = \frac{1}{1-x} \bigg|_{1/2}^{3/4}  = \int_{1/2}^{3/4}\frac{dx}{(1-x)^2}$$
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