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 $\sum_i x_i=1$ ?
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 $c_i$.