$\newcommand{\La}{\Lambda}$ The Legendre transform $\La^*_X$ of the log-moment generating function of a random variable $X$ is given by the formula $$\La^*_X(x):=\inf_{t\ge0}(-tx+\La_X(t)), $$ where $\La_X(t):=\ln Ee^{tX}$. The function $\La^*_X$ is the pointwise infimum in $t\ge0$ of the concave functions $x\mapsto-tx+\La_X(t)$. So, $\La^*_X$ is concave.
So, we may take $f_i=\La^*_i:=\La^*_{X_i}$ for all $i$. Then for $\La_i:=\La_{X_i}$ we have $$\La^*_{\sum_{i=1}^n X_i}(x)=\inf_{t\ge0}\sum_{i=1}^n(-t\tfrac xn+\La_i(t)) \ge\sum_{i=1}^n\inf_{t\ge0}(-t\tfrac xn+\La_i(t)) =\sum_{i=1}^n\La^*_i(\tfrac xn) =\sum_{i=1}^n f_i(\tfrac xn) \ge n\min_{i=1}^n f_i(\tfrac xn). $$ So\begin{align} \La^*_{\sum_{i=1}^n X_i}(x)&=\inf_{t\ge0}\sum_{i=1}^n(-t\tfrac xn+\La_i(t)) \\ &\ge\sum_{i=1}^n\inf_{t\ge0}(-t\tfrac xn+\La_i(t)) \\ &=\sum_{i=1}^n\La^*_i(\tfrac xn) \\ &=\sum_{i=1}^n f_i(\tfrac xn) \\ &\ge n\min_{i=1}^n f_i(\tfrac xn). \end{align}
So, we get the inequality opposite to what you suggested. Usually, this opposite inequality will be strict, if indeed the $X_i$'s are not identically distributed.