Any solution?

There is a really nice solution in Titu Andreescu's book, which goes as follows. I will just state idea, and will not carry out the details. Motivation Just like how we take the partition of a fixed, compact, interval, and compute upper/lower Darboux sums to evaluate Riemann integral. Suppose, these points are fixed, and consider the line $y=x$. The sum in the middle, can be interpreted as the sum of the areas of bunch of rectangles; while, the terms $1/2$, on left-hand-side, and right-hand-side, are the area of the triangle. The simple $h$ terms, are simply the departures from the triangle, and can be proven very easily.

It isn't difficult to solve it. We just need to consider the minimum and the maximum of all $x_{2k}$. On one hand, it's clear that $x_{2k}>x_{2k-1}$ and $x_{2k}\ge x_{2k+1}-h$ $\therefore x_{2k}\ge \max\{x_{2k-1},x_{2k+1}-h\}\ge \frac{1}{2}(x_{2k-1}+x_{2k+1}-h)$ $\therefore \sum\limits_{k=1}^nx_{2k}(x_{2k+1}-x_{2k-1})\ge \sum\limits_{k=1}^n\frac{1}{2}(x_{2k-1}+x_{2k+1}-h)(x_{2k+1}-x_{2k-1})=\frac{1}{2}\sum\limits_{k=1}^n\left(-(x_{2k+1}-x_{2k-1})h+(x_{2k+1}^2-x_{2k-1}^2)\right)=\frac{1-h}{2}$. On the one hand, we have $x_{2k}<x_{2k+1}$ and $x_{2k}\le x_{2k-1}+h$ $\therefore x_{2k}\le \min\{x_{2k-1},x_{2k+1}-h\}\le \frac{1}{2}(x_{2k-1}+x_{2k+1}+h)$ $\therefore \sum\limits_{k=1}^nx_{2k}(x_{2k+1}-x_{2k-1})\le \sum\limits_{k=1}^n\frac{1}{2}(x_{2k-1}+x_{2k+1}+h)(x_{2k+1}-x_{2k-1})=\frac{1}{2}\sum\limits_{k=1}^n\left((x_{2k+1}-x_{2k-1})h+(x_{2k+1}^2-x_{2k-1}^2)\right)=\frac{1+h}{2} $. Q.E.D