I would like to ask for a proof/disproof of a proposition.
Proposition 1. Suppose two functions
$f(x)=\sum_{i∈M}[a_i/(b_i+x)]$
and
$g(x)=\sum_{j∈N}[c_j/(d_j+x)]$
where $a_i,b_i,c_j,d_j>0$, $\min_{i∈M}${$b_i$} $>$ $\max_{j∈N}${$d_j$}, $x∈[x_1,x_2 ]⊆R$, $d_j+x>0$ $∀j∈N$. Let $h(x)=f(x)-g(x)$, then for any $x∈(x_1,x_2)$, there is $h(x)>$ min{$h(x_1),h(x_2)$}.
Thank you.
Following is a proof of a special case when all $b_i$/$d_i$ are identical.
Proposition 2. Suppose two functions
$f(x)=a/(b+x)$
and
$g(x)=c/(d+x)$
where $a,b,c,d>0$, $b>d$, $x∈[x_1,x_2 ]⊆R$, and $d+x>0$.
Let $h(x)=f(x)-g(x)$, then for any $x∈(x_1,x_2)$, there is $h(x)>$ min{$h(x_1),h(x_2)$}.
Proof. We use proof by contradiction to prove the proposition that
$h(x')≤h(x_1 )$ and $h(x')≤h(x_2 )$ $∃x'∈(x_1,x_2)$.
We have $h(x')≤h(x_1)$, or
$\frac{a}{b+x'}-\frac{c}{d+x'}≤\frac{a}{b+x_1}-\frac{c}{d+x_1}$.
By rearranging the inequality, we have $a\frac{(x_1-x')}{(b+x')(b+x_1)} ≤c\frac{(x_1-x')}{(d+x')(d+x_1)}$.
$∵a,c,x_1,x'>0$, $d+x>0$ and $x'>x_1$,
$∴a≥c\frac{(b+x')(b+x_1 )}{(d+x')(d+x_1)}$ (1)
Similarly, we have $h(x')≤h(x_2)$. Repeat the rearrangement above, we have
$a\frac{(x_2-x')}{(b+x')(b+x_2)} ≤c\frac{(x_2-x')}{(d+x_2)(d+x')}$.
$∵a,c,x_2,x'>0, d+x>0$ and $x_2>x'$,
$∴a≤c\frac{(b+x')(b+x_2)}{(d+x_2)(d+x')}$.
Together with Eq. (1), we have
$c\frac{(b+x')(b+x_1)}{(d+x')(d+x_1)}≤c\frac{(b+x')(b+x_2)}{(d+x_2)(d+x')}$.
By rearranging the inequality, we have
$\frac{b+x_1}{b+x_2}≤\frac{d+x_1}{d+x_2}⟹\frac{x_1-x_2}{b+x_2}≤\frac{x_1-x_2}{d+x_2}$ .
$∵x_1<x_2$, $∴b≤d$, which contradicts the initial conditions.
Edit: A slightly more complex version of Proposition 2 can also be proved using proof by contradiction. The proof is similar to the one above, interested readers may prove it by themselves.
Proposition 3. Suppose two functions
$f(x)=\sum_{i∈M}[a_i/(b_i+x)]$
and
$g(x)=c/(d+x)$
where $a_i,b_i,c,d>0$, $\min_{i∈M}${$b_i$} $>$ $d$, $x∈[x_1,x_2 ]⊆R$, $d+x>0$. Let $h(x)=f(x)-g(x)$, then for any $x∈(x_1,x_2)$, there is $h(x)>$ min{$h(x_1),h(x_2)$}.
I think I have the proof. Please see the screenshot below.