We define the monotone nonnegative cone as
$$K_{m+} = \{ x \in \mathbb{R}^n \mid x_1 \ge x_2 \ge \dots \ge x_n \ge 0\}$$
i.e. all nonnegative vectors with components sorted in nonincreasing order.
Find the dual cone $K^*_{m+}$. Hint: use the identity
\begin{align*} \sum^n_{i=1} x_iy_i = & (x_1 - x_2)y_1 + (x_2 - x_3)(y_1+y_2) + (x_3 - x_4)(y_1+y_2+y_3) \\ &+ \dots + (x_{n-1} -x_n)(y_1+ \dots + y_{n-1}) + x_n(y_1+ \dots + y_n) \end{align*}
I proved the cone is a proper cone if that knowledge helps somehow.
The same problem was posted here but the solution does not make sense to me.
The hint can be easily proven by induction on $n$, so I will skip the details for this. Assume $y\in K_{m+}^*$. For any given $k$ between $1$ and $n$, define $$ z_k = \begin{bmatrix} 1\\\vdots\\1\\0\\\vdots\\0 \end{bmatrix}, $$ where the last $1$ is in row $k$. Since $z_k\in K_{m+}$, we have $$ 0 \le y^Tz_k = y_1 + \cdots + y_k. $$ Reciprocally, if $y_1 +\cdots+ y_k \ge0$ for all $k$, the hint shows that $$ y^Tx = \sum_{i=1}^nx_iy_i $$ can be expressed as a sum of non-negative terms $(x_{i}-x_{i-1})(y_1+\cdots+y_i)$, which makes it non-negative, showing that $y\in K^*_{m+}$.