Hello Jutaro, Jutaro wrote: Given the positive real numbers $a_{1}<a_{2}<\cdots<a_{n}$, consider the function \[f(x)=\frac{a_{1}}{x+a_{1}}+\frac{a_{2}}{x+a_{2}}+\cdots+\frac{a_{n}}{x+a_{n}}\] Determine the sum of the lengths of the disjoint intervals formed by all the values of $x$ such that $f(x)>1$. Very nice question. Let $f(x)=\frac{P(x)}{Q(x)}$ with $P(x)=\sum_{i}a_{i}\prod_{j\neq i}(x+a_{j})$ and $Q(x)=\prod_{i}(x+a_{i})$ $f(x)=1$ $\Leftrightarrow$ $Q(x)-P(x)=0$ and this polynomial have exactly n roots $-a_{n}<r_{n}<-a_{n-1}<r_{n-1}<\cdots<-a_{1}<r_{1}$ The number requested is $\sum_{i}(r_{i}+a_{i})= \sum_{i}r_{i}+\sum_{i}a_{i}$ $\sum_{i}r_{i}$ is the opposite of the coefficient of $x^{n-1}$ in $Q(x)-P(x)$ Coefficient of $x^{n-1}$ in $Q(x)$ is $\sum_{i}a_{i}$ Coefficient of $x^{n-1}$ in $P(x)$ is $\sum_{i}a_{i}$ Hence $\sum_{i}r_{i}=0$ The the number requested is $\sum_{i}a_{i}$ -- Patrick

$f(x)-1=0\to \frac{Q(x)}{P(x)}=0$, were $P(x)=\prod_{i}(x+a_{i}), \ Q(x)=-P(x)+\sum_{i}a_{i}\frac{P(x)}{x+a_{i}}.$ Because $f(x)\to-\infty$, when $x\to a_{i}-0$, and $f(x)\to+\infty$, when $x\to a_{i}+0$ we have n roots Q(x)$-a_{n}<y_{n}<-a_{n-1}<...<-a_{1}<y_{1}<\infty$. We calculate ${Q(x)=-x^{n}-\sum_{i}a_{i}x^{n-1}-...+x^{n-1}(a_{1}+...+a_{n}}+...$. It give $\sum_{i}y_{i}=0$. Therefore $\sum_{i}L_{i}=\sum_{i}(y_{i}-(-a_{i}))=\sum_{i}a_{i}.$

pco wrote: ... The number requested is $\sum_{i}(r_{i}+a_{i})= \sum_{i}r_{i}+\sum_{i}a_{i}$ ... Sorry, but I'm having trouble understanding why it is this sum. I know I'm wrong, but I thought it would be something like $(r_1 - r_2) + (r_3 - r_4) + ...$ (that or $(r_2-r_3) + (r_4-r_5) + ... $). EDIT: Thanks pco!

modularmarc101 wrote: pco wrote: ... The number requested is $\sum_{i}(r_{i}+a_{i})= \sum_{i}r_{i}+\sum_{i}a_{i}$ ... Sorry, but I'm having trouble understanding why it is this sum. I know I'm wrong, but I thought it would be something like $(r_1 - r_2) + (r_3 - r_4) + ...$ (that or $(r_2-r_3) + (r_4-r_5) + ... $). No, you need to go back to the original expression of $f(x)$ : $f((-\infty,-a_n))=(-\infty,0)$ $f((-a_n,r_n))=(1,+\infty)$ and so we get a "good" interval $(-a_n,r_n)$ whose length is $r_n+a_n$ $f((r_n,-a_{n-1}))=(-\infty,1)$ $f((-a_{n-1},r_{n-1}))=(1,+\infty)$ and so we get another "good" interval $(-a_{n-1},r_{n-1})$ whose length is $r_{n-1}+a_{n-1}$ $f((r_{n-1},-a_{n-2}))=(-\infty,1)$ ... $f((-a_{n-k},r_{n-k}))=(1,+\infty)$ and so we get a "good" interval $(-a_{n-k},r_{n-k})$ whose length is $r_{n-k}+a_{n-k}$ $f((r_{n-k},-a_{n-k-1}))=(-\infty,1)$ Hence the result $\sum_{i}(r_{i}+a_{i})$