How to write dual form of $ \min \sum_{i=1}^{n} \mathbf{x}_i\mathbf{y}_i$ with $\sum_{i=1}^n \mathbf{y}_i = 1, \mathbf{y}_i \ge0$ $$\begin{array}{ll} \text{minimize}_{\mathbf{y} \in \mathbb R^n} &  \sum_{i=1}^{n} \mathbf{x}_i\mathbf{y}_i \\  \text{subject to} &  \ \sum_{i=1}^n \mathbf{y}_i = 1, \mathbf{y}_i \ge0
\end{array}$$
Here $\mathbf{x}, \mathbf{y}$ are both column vector with size $n$
How to write the dual form of this linear programming?

The Lagrangian function of the primal problem
$$
\begin{align*}
\min_{y \in \mathbb{R}^n}\; x^\top y \quad \text{s.t.} \quad \mathbb{1}^\top y = 1,\; y \geq 0 
\end{align*}
$$
is given by
$$
\begin{align*}
L(y, \lambda, \mu) = x^\top y + \lambda (1-\mathbb{1}^\top y) - \mu^\top y
= (x - \mathbb{1}\lambda - \mu)^\top y + \lambda.
\end{align*}
$$
The lagrangian dual problem reads
$$
\max_{\lambda \in \mathbb{R}, \mu \in \mathbb{R}^n } \inf_{y \in \mathbb{R}^n} L(y, \lambda, \mu) \quad \text{ s.t. } \quad \mu \geq 0.
$$
We have
$$
\begin{align*}
\inf_{y \in \mathbb{R}^n} L(y, \lambda, \mu) = \begin{cases} \lambda, \quad &{ x - \mathbb{1}\lambda - \mu = 0} \\ -\infty, \quad &\text{else} \end{cases}
\end{align*}
$$
and since $x - \mathbb{1}\lambda = \mu \geq 0$ iff $\mathbb{1}\lambda \leq x$, we have the dual problem
$$
\begin{align*}
\max_{\lambda \in \mathbb{R}} \; \lambda \quad \text{s.t.} \quad \lambda \leq x_i \; \forall i = 1, \ldots, n
\end{align*}
$$
