At most one positive root for a sum of fractions

Question

It seems that the following equation has at most one positive root $x \in \mathbb{R}$. How should I approach to prove it?

$\sum_{i=1}^K \frac{1}{x+id} = \sum_{i=1}^N \frac{1}{x+i}$

where $K < N$ are integers, and $d \in [0, 1]$.

Answer 1

Lemma 1: if $f_j,g_j>0$, $f_j(x)=g_j(x)$ and $f_j(y)/f_j(x)>g_j(y)/g_j(x)$ for all $j$, then $\sum_j f_j(y)>\sum_j g_j(y)$.

Proof: Subtract and enjoy.

Lemma 2: If $y>x>0$ and $0<a<1$, then $\frac{1+ax}{1+ay}>\frac{1+x}{1+y}$

Proof: Same as of Lemma 1.

First, multiply by $x$, and substitute $1/x$ for $x$. Next, let $x$ be the least positive root. Then $a_i\frac 1{1+xid}=\frac{1}{1+xi}$ for $i\le K$ with some $a_i\in (0,1)$ while the remaining parts of the sums can also be split as $b_{ij}\frac 1{1+xid}= c_{ij}\frac 1{1+xj}$ ($i\le K, K<j\le N$). Now apply the lemmas.