# Minimum of a sum of positive functions is the sum of the minimums of the functions

Let $$f_1, \dots, f_n$$ be positive functions from $$\mathbb R^m \rightarrow \mathbb R$$.

How do we show that $$\min_x \sum_{i=1}^n f_i(x) = \sum_{i=1}^n \min_x f_i(x)$$

Actually, I am not sure this is true. Maybe adding convexity of the functions helps ?

It is not true. Take $$f_1,f_2\colon\mathbb R\longrightarrow\mathbb R$$ defined by $$f_1(x)=(x-1)^2$$ and $$f_2(x)=(x+1)^2$$. Then $$\min f_1+\min f_2=0$$, but $$\min(f_1+f_2)=2$$.
• If each $f_i$ attains its minimum at a point $a$, then, for every $x\in\mathbb R^m$,$$\left(\sum_{i=1}^nf_i\right)(x)=\sum_{i=1}^nf_i(x)\geqslant\sum_{i=1}^nf_i(a)$$and therefore $\sum_{i=1}^nf_i$ attains its minimum at $a$. Jun 13, 2019 at 7:43