Infimum of this expression Let $n \in \mathbb{N}$ and $r_1, \dots, r_n$ be positive real numbers. Is it true that $$ \inf \left\lbrace r_1 x_1 + \cdots + r_n x_n : x = (x_1, \dots, x_n) \in \mathbb{R}^n, \, x_i \geq 0, \, \sum_{i=1}^n x_i^2 = 1 \right\rbrace = \min \\{ r_i : 1 \leq i \leq n\\}? $$ I suspect it is, and already showed that the infimum is smaller or equal to the minimum by simply considering the canonic vectors $e_i = (0, \dots, 1, \dots, 0)$. Can you help me to proof (or disprove) the reverse inequality?

Let $R = \min \\{r_i : 1 \le i \le n\\}$.

If each $x_i \ge 0$ and $\sum_{i=1}^n x_i^2 = 1$ then each $x_i$ satisfies $0 \le x_i \le 1$. In particular, $x_i^2 \le x_i$ for each index $i$. For any $n$-tuple of such $x_i$'s you get $$r_1x_1 + \cdots + r_nx_n \ge R (x_1 + \cdots + x_n) \ge R (x_1^2 + \cdots + x_n^2) = R.$$

Now take the infimum.