<s>[INST] 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.
 [/INST] Yes, there is a bound available.
Let $t_i := \max \Big\{ 
\tfrac{(x_i - \overline{x})^2}{x_i^3}, 
\tfrac{(x_i - \overline{x})^2}{\overline{x}^3} \Big\}$.
We have that the remainder is the average $\frac{1}{n} \sum_{j=1}^n t_j$.
Consider the two cases separately. First suppose $x_i > \overline{x}$.
Then
$$
t_i = \frac{(x_i - \overline{x})^2}{\overline{x}^3} \leq \frac{1}{\overline{x}} \frac{x_i^2}{\overline{x}^2} \leq (b/a)  \frac{1}{\overline{x}} \frac{x_i}{\overline{x}} 
$$
On the other hand, if $\overline{x} > x_i$,
$$
t_i = \frac{(x_i - \overline{x})^2}{x_i^3} \leq
\frac{1}{\overline{x}} \frac{\overline{x}^3}{x_i^3} \leq (b/a)^3 \frac{1}{\overline{x}}.  
$$
Therefore,
$$t_i \leq (b/a)^3 \frac{1}{\overline{x}} + (b/a)  \frac{1}{\overline{x}} \frac{x_i}{\overline{x}}$$.
If we average over $i=1, \dots, n$, we get
$$
\frac{1}{n}\sum_{i=1}^n f(x_i) \leq f(\overline{x}) + (b/a)^3 f(\overline{x}) + (b/a) f(\overline{x}), 
$$
so that we may take $C(a, b) = 1 + (b/a)^3 + (b/a)$.
</s>