How to prove $\sum_{i}a_i\alpha_i \geq \min_i(\alpha_i)$.

Question

How do I prove the following inequality: $$ \sum_{i=1}^N a_i \cdot \alpha_i \geq \min_{i} \{\alpha_i\}\,, $$ where $$ \sum_{i=1}^N a_i = 1\,, \text{and } a_i \geq 0\,. $$

Answer 1

You can write $$ \min_j \alpha_j = \min_j \alpha_j\sum_{i=1}^N a_i = \sum_{i=1}^N a_i \min_j \alpha_j \leq \sum_{i=1}^N a_i \alpha_i\,. $$ The inequality is because $\min_{1\leq j\leq N}\alpha _j \leq \alpha_i$ for all $i$.