Let $f(x)=a_{1} /\left(x+a_{1}\right)+a_{2} /\left(x+a_{2}\right)+\ldots+a_{n} /\left(x+a_{n}\right)$, where $a_{i}$ are unequal positive reals. Find the sum of the lengths of the intervals in which $f(x) \geq 1$.
Answer
$\sum a_{i}$
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```
wlog $a_{1}>a_{2}>\ldots>a_{n}$. The graph of each $a_{i} /\left(x+a_{i}\right)$ is a rectangular hyperbola with asymptotes $x=-a_{i}$ and $y=0$. So it is not hard to see that the graph of $f(x)$ is made up of $n+1$ strictly decreasing parts. For $\mathrm{x}-\mathrm{a}_{\mathrm{n}}, \mathrm{f}(\mathrm{x})$ decreases from $\infty$ to 0 . Thus $\mathrm{f}(\mathrm{x})=1$ at $\mathrm{n}$ values $\mathrm{b}_{1}<\mathrm{b}_{2}<\ldots<\mathrm{b}_{\mathrm{n}}$, and $f(x) \geq 1$ on the $n$ intervals $\left(-a_{1}, b_{1}\right),\left(-a_{2}, b_{2}\right), \ldots,\left(-a_{n}, b_{n}\right)$. So the sum of the lengths of these intervals is $\sum\left(a_{i}+b_{i}\right)$. We show that $\sum b_{i}=0$.

Multiplying $f(x)=1$ by $\prod\left(x+a_{j}\right)$ we get a polynomial of degree $n$ :

$$
\Pi\left(x+a_{j}\right)-\sum_{i}\left(a_{i} \prod_{j \neq i}\left(x+a_{j}\right)\right)=0
$$

The coefficient of $x^{n}$ is 1 and the coefficient of $x^{n-1}$ is $\sum a_{j}-\sum a_{i}=0$. Hence the sum of the roots, which is $\sum b_{i}$, is zero.
```