Relying only on the number of variables

Since the numerator is $1-x_n$ , we have $f(x_1,x_2,\ldots,x_n)=-n+\sum \frac{1}{x_i}$

To minimise $f$ , we need to minimise $\sum \frac{1}{x_i}$ subject to $\sum x_i=1$ .

But this is just the AM-HM inequality: $\begin{aligned} \frac{\sum x_i}{n} &\ge \frac{n}{\sum \frac{1}{x_i}} \\ \frac{1}{n} &\ge \frac{n}{\sum \frac{1}{x_i}} \\ \sum \frac{1}{x_i} &\ge n^2 \end{aligned}$

Hence the minimum value of $f$ is $n^2-n$ , or $\boxed{90}$ in this case. (This minimum is achieved when $x_i=\frac{1}{n}$ for each $i$ .)

For positive real numbers $x_1, x_2, \dots, x_{10}$ , the given sum may be written as $\left( \dfrac{1}{x_1} + \dfrac{1}{x_2} + \cdots + \dfrac{1}{x_{10}}\right) \left( x_1 + x_2 + \cdots + x_{10}\right) - \dfrac{x_1}{x_1} - \dfrac{x_2}{x_2} - \cdots - \dfrac{x_{10}}{x_{10}}$ .

Hence, by the Cauchy-Schwarz inequality, we see that $\displaystyle f(x_1, x_2, \dots, x_{10}) \ge \left( \sum_{i=1}^{10} \dfrac{x_i}{x_i}\right)^2 - 10 = 10^2 - 10 = 90.$