The polynomial has a root in the interval $(a_1,a_1+1).$ Let $a_1,a_2,\ldots,a_k$ and $p$ are positive real numbers where $a_1$ is minimum. If we solve $$\frac{1}{x-a_1}+\frac{1}{x-a_2}+\ldots \frac{1}{x-a_k}-1=0 \quad(1)$$ we have a $k$-degree polynomial.
Consider another polynomial, given by: $$\frac{1}{x-a_1}+\frac{1}{x-a_2}+\ldots \frac{1}{x-a_k}-\frac{p}{x}-1=0\quad(2)$$
I have two observation:  

$(a)$ Least root of $(1)$ lies in $(a_1,a_1+1).$
  $(b)$ Least root of $(2)$ is less than that of $(1).$  

Are the two observations true? If so, how to prove them?

Here is part of (a):
If $x < a_1 \le a_i$, then $x-a_i <0$ and so $\displaystyle\frac{1}{x-a_1}+\frac{1}{x-a_2}+\cdots \frac{1}{x-a_k} < 0 < 1$.
Therefore,  there is no root smaller than $a_1$.
