I am trying to prove the following statement.
given vectors $\textbf{a},\textbf{a}_1, \cdots, \textbf{a}_m \in \mathbb{R}^n$,
$\forall \textbf{x} \geq 0 $, we have $\textbf{a}'x \leq \textrm{max}_i \{ {\textbf{a}'_i \textbf{x}}\}$ if and only if there exist nonnegative coefficients $\lambda_i$ that sum to 1 and such that $\textbf{a} \leq \sum_{i=1}^m\lambda_i\textbf{a}_i$
here $\textbf{v} \leq \textbf{w}$ for $\textbf{v}, \textbf{w}\in \mathbb{R}^n$ means we have $v_i \leq w_i$, $i = 1, \cdots, n$
for one way, it was pretty straightforward,
$\textbf{a}'\textbf{x} \leq \sum_{i=1}^m\lambda_i\textbf{a}_i'\textbf{x} \leq \sum_{i=1}^m\lambda_i\textrm{max}_j\{\textbf{a}_j'\textbf{x}\} \leq \textrm{max}_j\{\textbf{a}_j'\textbf{x}\}\sum_{i=1}^m\lambda_i \leq \textrm{max}_j\{\textbf{a}_j'\textbf{x}\}$
showing the other way is much of a struggle for me and I dont really know where to start. I have thought of trying maybe forming a polyhedron thats convex hull of $\textbf{a}_1, \cdots, \textbf{a}_m$ but it brought me nowhere. I have also attempted maybe forming a cone but it seems $\textbf{a}$ can be lying outside the cone. Any hint or guidance is appreciated.
To show (a)$\implies$(b), we can show $\neg$(b)$\implies$$\neg$(a). To begin, assume that for every $\lambda\geq0$ with $\lambda^\top \mathbf{1}=1$ that $a \not\leq \sum_{i=1}^m\lambda_i a_i$. Then there is a coordinate $k\in\{1,2,\ldots,n\}$ of $a$ such that $a^{k}>\sum_{i=1}^m\lambda_i a_i^k$. Next consider a special $x:=e_k=(0,\ldots,1,\ldots,0)$, the $k$th standard basis vector. Taking a special $\lambda^*$ to be the vector of zeros with a one in the $i^*$th slot where $i^* := \mathrm{arg}\max_i \{a_i^\top x\}$, we have $$a^\top x = a^k > \sum_{i=1}^m\lambda_i^* a_i^k = \sum_{i=1}^m\lambda_i^* a_i^\top x = \max_i\{a_i^\top x\}$$ and are done.