We will reason by induction on $\mathrm{n} \geqslant 1$.

- For $n=1$, consider two real numbers $x_{1}, y_{1} \in [0,1]$ and set $x_{2}=1$. We need to prove that $\left(x_{1}-y_{1}\right)+\left(x_{2}-x_{1}\right) y_{1} \geqslant 0$.

Indeed, we have $\left(x_{1}-y_{1}\right)+\left(x_{2}-x_{1}\right) y_{1}=\left(x_{1}-y_{1}\right)+\left(1-x_{1}\right) y_{1}=x_{1}\left(1-y_{1}\right) \geqslant 0$, which concludes the proof.

- Suppose that for some $n \geqslant 1$ and for all real numbers $0 \leqslant a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant 1$ and $0 \leqslant b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n} \leqslant 1$ with $a_{n+1}=1$, we have

$$\sum_{i=1}^{n}\left(a_{i}-b_{i}\right)+n \sum_{i=1}^{n}\left(a_{i+1}-a_{i}\right) b_{i} \geqslant 0$$

Consider then real numbers $0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n+1} \leqslant 1$ and $0 \leqslant y_{1} \leqslant y_{2} \leqslant \cdots \leqslant y_{n+1} \leqslant 1$ and set $x_{n+2}=1$.

By isolating the contributions of $x_{1}$ and $y_{1}$, we have

$$\begin{aligned} & \sum_{i=1}^{n+1}\left(x_{i}-y_{i}\right)+(n+1) \sum_{i=1}^{n+1}\left(x_{i+1}-x_{i}\right) y_{i} \\ &= \sum_{i=2}^{n+1}\left(x_{i}-y_{i}\right)+n \sum_{i=2}^{n+1}\left(x_{i+1}-x_{i}\right) y_{i}+x_{1}-y_{1}+\sum_{i=2}^{n+1}\left(x_{i+1}-x_{i}\right) y_{i} \\ & \quad+(n+1)\left(x_{2}-x_{1}\right) y_{1} \end{aligned}$$

applied to the real numbers $a_{i}=x_{i+1}$ and $b_{i}=y_{i+1}$

$$\geqslant x_{1}-y_{1}+\sum_{i=2}^{n+1}\left(x_{i+1}-x_{i}\right) y_{1}+(n+1)\left(x_{2}-x_{1}\right) y_{1} \text{ since } y_{1} \leqslant y_{i} \text{ and } x_{i+1} \geqslant x_{i} \text{ for }$$

all $i$

$$\begin{aligned} & =x_{1}+y_{1}\left[-1+(n+1)\left(x_{2}-x_{1}\right)+\sum_{i=2}^{n+1}\left(x_{i+1}-x_{i}\right)\right] \\ & =x_{1}+y_{1}\left[n x_{2}-(n+1) x_{1}\right] \\ & =x_{1}\left(1-y_{1}\right)+n y_{1}\left(x_{2}-x_{1}\right) \\ & \geqslant 0 \end{aligned}$$

which proves the desired result for the value $n+1$ and completes the proof.
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