Converted from my comments. The first observation is that you can assume the $y_i$ to be nonnegative scalars: indeed, you can replace $y_i$ with $\|y_i\|$ without changing the problem.

Then, the thesis follows from the triangle inequality. Indeed,
$$
\begin{aligned}
\|z_1\| + \|z_2\| + \dots + \|z_n\| &\geq \|z_1 + z_2 + \dots + z_n\| \\ & = \left\|\begin{bmatrix}c \\ \overline{y}\end{bmatrix}\right\| \\&= \left\|\begin{bmatrix}c\frac{y_1}{\overline{y}} \\ y_1\end{bmatrix} + \dots + \begin{bmatrix}c\frac{y_n}{\overline{y}} \\ y_n\end{bmatrix} \right\|
\\&= \left\|\begin{bmatrix}c\frac{y_1}{\overline{y}} \\ y_1\end{bmatrix}\right\| + \dots + \left\|\begin{bmatrix}c\frac{y_n}{\overline{y}} \\ y_n\end{bmatrix} \right\|.
\end{aligned}
$$
with $\overline{y} = y_1 + y_2 + \dots + y_n$. The last equality holds because all the vectors in the last expression are nonnegative scalar multiples of $\begin{bmatrix}c\frac{1}{\overline{y}} \\ 1\end{bmatrix}$.

(If $\overline{y}=0$, then you can take any partition of $1$ in place of the quantiieis $\frac{y_i}{\overline{y}}$ and the proof works in the same way.)

The geometric idea behind this proof is the following: let $k=j-k=1$ for simplicity, so that we are on the plane. You wish to travel from the point $(0,0)$ to $(c,\overline{y})$, and you are allowed to move along a polygonal line that bends only in points with ordinates $y_1, y_1+y_2,\dots, y_1+y_2+\dots+y_{n-1}$. Then, clearly, the shortest path is a straight line, making no bends.

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/JAcS9.png