A lower bound to $\sum_{i=1}^n \frac{1}{a + x_i}$ with given variance What is a tight lower bound to $\sum_{i=1}^n \frac{1}{a + x_i}$ under the restrictions  $\sum_{i=1}^n x_i = 0$ and $\sum_{i=1}^n x_i^2 = a^2$ ?
Conjecture: due to the steeper rise of  $\frac{1}{a + x}$ for negative $x$, one may keep those values as small as possible. So take $n-1$ values $x_i = -q$ and $x_n = (n-1) q$ to compensate for the first condition. The second one then gives  $a^2 = \sum_{i=1}^n x_i^2 = q^2 ((n-1)^2 + n-1) = q^2 n (n-1)$. Hence,
$$\sum_{i=1}^n \frac{1}{a + x_i} \ge \frac{n-1}{a (1- 1/\sqrt{n(n-1)})} + \frac{1}{a (1+ (n-1)/\sqrt{n(n-1)})}$$ should be  the tight lower bound.
 A: Yes, this is indeed the minimum value (assuming $a>0$).
Denote $K=\{x\in\mathbb{R}^n\mid\sum_{k=1}^n x_k=0,\sum_{k=1}^n x_k^2=a^2\}$ and let $f(x)=\sum_{k=1}^n(a+x_k)^{-1}$ attain its minimum at $x=\bar{x}\in K$ (it does so, as a continuous function on $\{x\in K\mid f(x)\leqslant f(0)\}$ which is compact). Then, by Lagrange multiplier theorem, there are $\lambda_1,\lambda_2\in\mathbb{R}$ such that for each $k$ we
have $(a+\bar{x}_k)^{-2}=\lambda_1+\lambda_2\bar{x}_k$. Then the positive numbers $y_k=a+\bar{x}_k$ are solutions of $$y^2(\lambda_1-\lambda_2 a+\lambda_2 y)=1.$$ But this equation has at most two positive solutions. Thus, at most two values among $\bar{x}_k$ are distinct, and in fact exactly two. So, let $m$ values of $\bar{x}_k$ equal $b>0$, where $0<m<n$, and the remaining $n-m$ values equal $c<0$. We get a system for $b$ and $c$, solve it, and finally obtain $$af(\bar{x})=n+\left(1-\frac1n+\frac{n-2m}{\sqrt{nm(n-m)}}\right)^{-1}.$$ The least possible value of this is at $m=1$.
