Lagrange optimization of reciprocals Suppose we want to find the $x_i$ such that $$\sum_{i=1}^p\frac{1}{1+x_i}$$
has an extreme value, such that the $x_i$ are all non-negative, and $\sum_{i=1}^n x_i=p$.
This should be doable with lagrange multipliers, but how? The inequality is tricky, so I considered replacing the $x_i$ with $x_i^2$. However, each entry of the gradient gives you $(1+x_i^2)^{-2}2x_i=\lambda 2x_i$, so I'm not quite sure how to proceed. It seems obvious that you get a min when all the $x_i$ are 1, but what about the maximum?
Edit: It was noted that I need to check the boundary points for the maximum. Is there a systematic technique to do this? I suspect that the max will be attained when all but one of the $x_i$ is zero, but how do you start when the boundary contains so many points?
 A: Let
$$ L = \sum_{i=1}^n \frac{1}{1+x_i} - \lambda \left( \sum_{i=1}^n x_i -p \right). $$
Then
$$ \frac{\partial L}{\partial x_j} = \frac{-1}{(1+x_j)^2}-\lambda, $$
and so we discover that if the gradient of $L$ is zero (required for an internal extremum), all of the $x_j$ are equal, and in particular, equal to $p/n$, by the constraint. In this case, the value of $L$ is
$$ \frac{n}{1+p/n} = \frac{n^2}{n+p} = \frac{n(n+p)-np-p^2+p^2}{n+p} = n-p + \frac{p^2}{n+p} $$.
What if we don't have this case? Then the extremum occurs on the boundary, i.e. when one of the $x_i$ at least is zero. By symmetry we can assume $x_n=0$, and $L$ becomes
$$ L\mid_{x_n=0} = 1 + \sum_{i=1}^{n-1} \frac{1}{1+x_i} - \lambda \left( \sum_{i=1}^{n-1} x_i - p \right). $$
Ah, but this is exactly the same problem as we solved before, but with $n$ replaced by $n-1$. Hence the extremum is at $x_i=p/(n-1)$, and so the value is
$$ 1+ \frac{(n-1)^2}{n-1+p} = \frac{n-1+p-(n-1)+n(n-1)}{n-1+p} = n-p+\frac{p^2}{n-1+p}. $$
Similarly, if $k$ of the $x_i$ are zero, we find
$$ k+\frac{(n-k)^2}{n-k+p} = k+n-k-p + \frac{p^2}{n-k+p} = n-p + \frac{p^2}{n-k+p}, $$
which obviously increases as $k$ increases. Hence the global maximum is when $k=n-1$, with value $n-1+\frac{1}{1+p}$, whereas the global minimum is when all the $x_i$ are $p/n$ (can check this is a minimum using the Hessian), with value $\frac{n}{1+p/n}$.
