Optimisation with unit simplex minimise $$\sum_i x_ic_i$$ for constant $c \in \mathbb{R}^n$, subject to $\sum_i x_i=1$ as $x_i \geq 0$

we are told to use Lagrange's method, so I have $$L(x,\lambda)=\sum_i x_ic_i 
-\lambda(\sum_i x_i-1)$$
So $$\frac{\partial L}{\partial x_i}=c_i -\lambda =0$$
But this gives no dependency for each individual $x_i$ ?
What are the general approaches for solving convex optimisation problems with a probability distribution such as $\sum_i x_i=1$ ?
 A: First of all, "Lagrange's method" with inequality constraints is a little bit different, is called the KKT Conditions, and includes a bit more than setting the derivative to zero. The Lagrangian is
$$
L(x, \mu, \lambda) = \sum_{i=1}^n c_i x_i + \mu(\sum_{i=1}^n x_i - 1) +\sum_{i=1}^n \lambda_i (-x_i),
$$
with $\lambda_i \geq 0$, and the conditions are
$$
\begin{aligned}
c_i + \mu - \lambda_i &= 0 & \leftarrow\nabla L = 0 \\
\lambda_i (-x_i) &= 0 & \leftarrow\text{Complementarity} \\
\sum_{i=1}^n x_i = 1,~ x &\geq 0 & \leftarrow\text{Primal feasibility} \\
\lambda &\geq 0 &\leftarrow\text{Multiplier (dual) feasibility}
\end{aligned}
$$
Unfortunately, it is hard to find a vector which satisfies these conditions directly. Personally, I am not aware of an elegant way to do so.
However, another approach can be used which does not involve Lagrange's method at all.  The unit simplex is a convex set, and the problem is equivalent to maximizing $\sum_{i=1}^n (-c_i x_i)$, which is also convex. It is known that the maximum of a convex function on a convex set is attained at an extreme point. In the case of the simplex, these are its vertices - the standard basis vectors $\mathrm{e}_i = (0, \dots, 1, \dots, 0)^T$.
Thus, after casting back to the minimization form, the minimum can be computed by
$$
\min_{i=1, \dots, n} \{ {\mathrm{e}_i}^T x \} = \min_{i=1, \dots, n} c_i 
$$
and the optimal solution is the corresponding vector $x^* = \mathrm{e}_{i^*}$, where $i^*$ is the index of one of a minimum $c_i$.
