How to prove the monotonicity of the following function? $f(x) = \frac{\sum\limits_{i=1}^K\ln(1+a_ix)}{\sum\limits_{i=1}^K\ln(1+\frac{1}{2}a_ix)}$ How to prove the monotonicity of the following function? 
$f(x) = \frac{\sum\limits_{i=1}^K\ln(1+a_ix)}{\sum\limits_{i=1}^K\ln(1+\frac{1}{2}a_ix)}$.
where $a_i>0$, $\forall i$, $x>0$.
I have proved that $\frac{h_i(x)}{g_i(x)}=\frac{\ln(1+a_ix)}{\ln(1+\frac{1}{2}a_ix)}$ is a monotonically decreasing function with $x$. Actually, $f(x)$ is also monotonically decreasing with $x$, which can be proved by simple simulation. But I can't give the rigorous proof. 
Any help is appreciated. Thanks a lot!
 A: OK, I put the remark yesterday at night just to prevent people from wasting time on attempts to prove the statement. Now it is time to post the explanation. Note that if $\frac{F(x)}{G(x)}$ is non-increasing and $F,G>0$, then $\frac{F'(x)}{F(x)}\le \frac{G'(x)}{G(x)}$ and, in particular, if $G'(x)-cG(x)\le 0$ for some positive $c$ and $x$, then $F'(x)-cF(x)\le 0$.
Now let us look at what functions we have. Repeating the summands, if necessary, and approximating reals by rationals, we can put $F(x)=\int\log (1+2ax)\,d\mu(a)$ and $G(x)=\int\log(1+ax)\,d\mu(a)$ where $\mu$ is any finite sum of Dirac point masses with positive coefficients. So, if our inequality holds, we must have that $\int \left[\frac{a}{1+ax}-c\log(1+ax)\right]\,d\mu(a)\le 0$ implies that $\int \left[\frac{2a}{1+2ax}-c\log(1+2ax)\right]\,d\mu(a)\le 0$ for every fixed $x,c>0$ and every $\mu$ of the above kind. We'll take $x=1$, $c\in(0,1)$.
Then, by the general abstract duality nonsense, for any finite set $A$ of points $a>0$, we must be able to produce a non-negative $\lambda$ such that
$$
\frac{2a}{1+2a}-c\log(1+2a)\le\lambda\left[\frac{a}{1+a}-c\log(1+a)\right]
$$
for all $a\in A$. Now just take $A$ consisting of $2$ points: $a_1\approx 0$ and $a_2\approx+\infty$.
Looking at $a_1$, we get essentially $2a_1(1-c)\le \lambda a_1(1-c)$, so $\lambda$ cannot be noticeably less than $2$. Looking at $a_2$, we get essentially $-c\log a_2\le \lambda(-c\log a_2)$, so $\lambda$ cannot be noticeably greater, than $1$. But this leaves us no room! 
