Minimize a linear function over the unit ball. (without using Lagrangian)
We have a problem: minimize $c^Ty$ subject to $y^Ty \le 1$.
By the Cauchy-Schwarz inequality we have $c^Ty \le \|c\|_2\|y\|_2 \le \|c\|_2$. So, if the problem was to maximize we would get $c^Ty = \|c\|_2, y= c/\|c\|_2$.
The answer to the problem is $y^{*} = -c/\|c\|_2$. However, I don't see how it can be derived. One thought is:
The problem is equivalent to: maximize $-c^Ty = (-c)^Ty$ subject to $y^Ty \le 1$.
$(-c)^Ty \le \|c\|_2, y = -c/\|c\|_2$
You can write the Lagrangian of the problem:
$\mathcal{L}(y, \lambda) = c^T y - \lambda(y^Ty - 1)$ with $\lambda$ the dual variable associated with the constraint.
The optimality conditions are:
$\frac{\partial \mathcal{L}}{\partial y} = c - 2 \lambda y = 0$
$y^Ty \le 1$
$\lambda(y^Ty - 1) = 0$
$\lambda \le 0$
We must have $\lambda < 0$ (constraint is active at the solution: $y^T y = 1$), so $y = \frac{c}{2 \lambda}$, which we inject in the 3rd equation, solve for $\lambda$, substitute in $y$ and we get the result.