Converse to Jensen's inequality for $1/x$ on a positive bounded interval? Consider the function $f(x) = 1/x$ on the interval $I = [a, b]$, where $0 < a \leq b$.
By Jensen's inequality, we have for any $\{x_j \}_{j=1}^n \subset I$,
$$
f(\overline{x}) \leq \frac{1}{n} \sum_{i=1}^n f(x_i). 
$$
Above, $\overline{x} = (1/n)\sum_{j=1}^n x_j$.
Is there a converse inequality of the form
$$
\frac{1}{n} \sum_{i=1}^n f(x_i) \leq C f\Big(\frac{1}{n}\sum_{i=1}^n x_i\Big), 
$$
where $C$ may depend on $a, b$?
One thing I thought of is a second-order Taylor expansion.
By Taylor expansion, there are $\xi_i$ in the interval between $x_i, \overline{x}$ such that
$$
\frac{1}{n} \sum_i f(x_i) = f(\overline{x}) + \frac{1}{2n} \sum_{i=1}^n f''(\xi_i)
(x_i - \overline{x})^2. 
$$
Additionally, we have that $f''(\xi) = \tfrac{2}{\xi^3}$, so that the remainder is
$$
\frac{1}{n} \sum_{i=1}^n \frac{(x_i - \overline{x})^2}{\xi_i^3} 
\leq 
\frac{1}{n} \sum_{i=1}^n \max \Big\{ 
\frac{(x_i - \overline{x})^2}{x_i^3}, 
\frac{(x_i - \overline{x})^2}{\overline{x}^3} \Big\}.
$$
I struggled to bound this remainder in terms of $C(a, b)~f(\overline{x})$, though.

We want to determine an upper bound for the function
$$
 h(x_1, \ldots, x_n) = \frac{1}{n^2} \sum_{k=1}^n x_k\sum_{k=1}^n \frac{1}{x_k}
= \sum_{k, l=1}^n \frac{x_k}{x_l}
$$
on the hypercube $[a, b]^n$. $h$ is convex in each variable, so that the maximum of $h$ is attained at a corner point of the hypercube. Therefore it suffices to compute
$$
\begin{align}
 h(\underbrace{a, \ldots, a}_{m},\underbrace{b, \ldots, b}_{n-m})
&= \frac{1}{n^2}\bigl(ma - (n-m)b\bigr)\left( \frac m a + \frac{n-m} {b}\right) \\
&= \left( \frac mn \right)^2 + \frac mn\left( 1-\frac mn \right)\left( \frac a b + \frac b a\right) + \left( 1-\frac mn \right)^2 \\
&= \dots \\
&= \frac{(a+b)^2}{4ab} - \frac{(a-b)^2}{4ab}\left(2 \frac mn - 1\right)^2 \, .
\end{align}
$$
The sharp upper bounds are therefore
$$
 h(x_1, \ldots, x_n) \le \frac{(a+b)^2}{4ab}
$$
if $n$ is even, and
$$
 h(x_1, \ldots, x_n) \le \frac{(a+b)^2}{4ab} - \frac{1}{n^2}\frac{(a-b)^2}{4ab}
$$
if $n$ is odd.
The best upper bound which is independent of $n$ is $C = \frac{(a+b)^2}{4ab}$.
