Strongly convex function evaluated over a mean of n points Let f(x) be a Strongly-convex function under some m > 0.
Given two points x, y it is known that:
$$f(\frac{x + y}{2}) \leq \frac{f(x) + f(y)}{2} - \frac{1}{2^3} \cdot m \cdot ||x - y||^2$$
What is the exact formulation when looking at the mean of n points.
$$f( \frac{1}{n}\sum_{i=1}^{n} x_i) \leq \ \  ?$$
 A: The standard definition of strong convexity with a constant $m>0$ would read as follows: for $t \in [0,1]$,
                  $$ f(tx+(1-t)y) \leq t f(x) + (1-t) f(y) - mt(1-t) \left \| x-y  \right \|^2.$$
Now, there is a famous theorem that says that a function $f$ is strongly convex with constant $m>0$ if and only if the function $$g(x)=f(x)-m\left \| x \right \|^2$$ is convex.
By Jensen's inequality for convex functions, $$ g\Bigg(\frac{\sum_{i=1}^{n} x_i }{n} \Bigg)  \leq \frac{1}{n} \sum_{i=1}^{n} g(x_i).$$ Equivalently, 
$$ f\Bigg(\frac{\sum_{i=1}^{n} x_i }{n} \Bigg)   \leq m \left \| \frac{\sum_{i=1}^{n} x_i }{n} \right \|^2 + \frac{1}{n} \sum_{i=1}^{n} \Big(f(x_i)- m \left \| x_i\right \|^2\Big) \\ \leq \frac{m(1-n)}{n^2} \sum_{i=1}^{n} \left \| x_i\right \|^2 + \frac{1}{n} \sum_{i=1}^{n} f(x_i),$$ 
where the last step follows from the triangle inequality.
See for example 

Nikodem, K. and Pales, Z., (2011). Characterizations of inner product spaces by strongly convex functions, Banach Journal of Mathematical Analysis, 5.1, 83-87.

for reference.
