prove that $\phi(a)=\frac{\int_{0}^{0.5} (\frac{u}{1-u})^{2a-1} du}{\int_{0.5}^{1} (\frac{u}{1-u})^{2a-1} du}>1 \Longleftrightarrow a<0.5$. Let $a\in (0,1) $, the  question is how to prove
$$\phi(a)=\frac{\int_{0}^{0.5} (\frac{u}{1-u})^{2a-1} du}{\int_{0.5}^{1} (\frac{u}{1-u})^{2a-1} du}>1 \Longleftrightarrow a<0.5$$
The plot of $(a,\phi (a))$ shows $\phi$ is monotone.

R Code
a<<-.5
fu<-function(u){
ret.value<-((u)/(1-u))^(2*a-1)
return(ret.value)
}
#(integrate1<-integrate(fu,lower=0,upper=.5)$value)
#(integrate2<-integrate(fu,lower=.5,upper=1)$value)

s<-seq(.1,.9,len=100)
ratio<-c()
for(i in 1:length(s)){
a<<-s[i]
ratio[i]<-integrate(fu,lower=0,upper=.5)$value/integrate(fu,lower=.5,upper=1)$value
}
plot(s,ratio,typ="l",axes=F,xlim=c(0,1))
axis(1,pos=0, (0:4)/4,(0:4)/4, col.axis = "black",padj=-.7,
lwd.ticks=1 ,tck=-.01,cex.axis=.95)
axis(2,pos=0, c(0,1,5,10,15,20),c("",1,5,10,15,""), col.axis = "black",padj=.4,
lwd.ticks=2 ,tcl=-.1,las=1, hadj=.4,cex.axis=.95)
abline(v=0.5,col="red")
abline(h=1,col="red")

 A: U can see that $u/(1-u)$ is a strictly increasing function from 0 to 1 at 0 it is 0 and at 1 it is infinite so when u are doing that integration the one with limits 0.5 to 1 will be of very greater value compared to that from 0 to 0.5 so to get numerator as higher value you should have power of negative value so that lower becomes higher and vice versa
A: Let us consider the integral at denominator. With the change of variable $t=1-u$ it can be rewritten as
$$
\int_0^{0.5} \left(\frac{1-t}{t}\right)^{2a-1}dt = \int_0^{0.5}\left(\frac{t}{1-t}\right)^{1-2a}\, dt,
$$
so that
$$
\phi(a) = \frac{\int_0^{0.5} f(u)^{2a-1}\, du}{\int_0^{0.5} f(u)^{1-2a}\, du}\,,
$$
with $f(u) = u/(1-u)$.
It is easy to check that $0 < f(u) < 1$ for every $u \in (0, 0.5)$.
Hence, if $2a - 1 < 0$, i.e. if $a < 0.5$, we have that
$$
f(u)^{2a-1} > f(u)^{1-2a}, \quad \forall u \in (0, 0.5)
\qquad (a < 0.5),
$$
so that $\phi(a) > 1$.
On the other hand, the inequality is reversed if $a > 0.5$ and, finally, $\phi(a) = 1$ if $a = 0.5$.
