For $x,y\in\mathbb{R}^d$, we say $x\succeq y$ if
$$\sum_{i=1}^j x_i\geq\sum_{i=1}^j y_i \text{ for all }1\leq j< d,\text{ and }\sum_{i=1}^d x_i=\sum_{i=1}^d y_i.$$
If $x\succeq y$, define $$\mathcal{T}(x,y):=\sum_{i=1}^d i(y_i-x_i)=\sum_{j=1}^d\sum_{i=1}^j(x_i-y_j).$$ Showing the two expressions are the same is not bad.
Claim: $$\lVert x-y\rVert_1\leq\mathcal{T}(x,y),$$ where $\lVert\cdot\rVert_1$ is the Taxicab norm.
One proof (found in the paper "The Pasulsen Problem Made Simple" by Hamilton and Moitra) shows that $\mathcal{T}$ is equivalent to the Wasserstein distance on the real line using its dual version. Then the inequality becomes a comparison of costs of moving piles in different ways.
Is there a more elementary way to prove this? I only need the result for nonnegative vectors $x$ and $y$, i.e., all their coordinates are nonnegative.