prove $\max \mathbf{a}^T \mathbf{b}_i \leq \mathbf{a}^T \mathbf{c}_i$

Question

Can I prove the following (with or without assumptions, e.g. all the elements in $\mathbf{a}$ or $\mathbf{b}$ are positive?

$\max \mathbf{a}^T \mathbf{b}_i \leq \mathbf{a}^T \mathbf{c}_i$

where $\mathbf{a}$ and $\mathbf{b}_i$ are vectors of dimension $d$.

Let $B = \left[\mathbf{b}_1, \dots, \mathbf{b}_N\right]$, $\max{c}_i$ represents a vector which has the rowwise maximum ofthe matrix B.

e.g.

e.g. $B = \begin{bmatrix} 1 & 2 & 4 \\ 5 & 1 & 2 \\ 1 & 4 & 1 \end{bmatrix} \implies \mathbf{c}_i = \begin{bmatrix} 4 \\ 5 \\ 4 \end{bmatrix}$

Answer 1

The elements in $a$ must be non-negative. (Take $a=-1$ for a start of a counterexample)

If $a\ge 0$ then $$ a^Tb_i = \sum_{j=1}^d a_j b_{i,j} \le \sum_{j=1}^d a_j \max_i b_{i,j}= a^T (\max b_i). $$ Now take the maximum on the left-hand side.