Inequality, Vojtěch Jarník Competition 2006 This is the problem from Vojtěch Jarník Competition 2006. Given real numbers $0=x_1,x_2<\dots<x_{2n}<x_{2n+1}=1$ such that $x_{i+1}-x_{i}\leq h$ for $1\leq i \leq 2n$, show that
$$\frac{1-h}{2}<\sum^n_{i=1}x_{2i}(x_{2i+1}-x_{2i-1})<\frac{1+h}{2}$$
So far I have achieved some results, but couldn't proceed any further:
$$\sum^n_{i=1}x_{2i}(x_{2i+1}-x_{2i-1})=x_2(x_3-x_1)+x_4(x_5-x_3)+x_6(x_7-x_5)+\dots=x_3(x_2-x_4)+x_5(x_4-x_6)+\dots$$

$$x_{i+1}-x_{i}\leq h$$
$$x_{i}-x_{i-1}\leq h$$
$$\dots$$
$$\frac{x_{i+1}-x_{i-m}}{m+1}\leq h$$

$$x_{2i-1}>\frac{x_{2i-1}+x_{2i-2}}{2}>x_{2i-1}x_{2i-2}>x_{2i-2}$$
$$x_{2i}>\frac{x_{2i}+x_{2i-1}}{2}>x_{2i}x_{2i-1}>x_{2i-1}$$
$$x_{2i-1}-x_{2i}>\frac{x_{2i-2}-x_{2i}}{2}>x_{2i-1}(x_{2i-2}-x_{2i})>x_{2i-2}-x_{2i-1}\geq -h$$

$$\frac{1+h}{2}\geq \frac{x_i-x_{i-n}+n}{2n}=\dfrac{\dfrac{x_i-x_{i-n}}{n}+1}{2}>\frac{x_i-x_{i-n}}{n}$$
I would not mind to see not only hints but full proofs as well, in case I am nowhere near the truth
 A: The sum in question is a Riemann sum approximating $\int_0^1x\,dx$. But you don't need to know that. Draw a picture and you see the sum is the area of a bunch of rectangles. 

The area of the rectangles equals $T + P - N$, where $T=1/2$ is the area of the triangle with vertices $(0,0)$, $(1,0)$ and $(1,1)$, $P$ is the green area, and $N$ is the red area. Each of the red triangles is confined to its own parallelogram (from $x_{2i-1}$ to $x_{2i+1}$, with height $h$) so the total red area is at most $h/2$. Similarly the green area is at most $h/2$.  The result follows from $T-N\le T+P-N\le T+P$.
A: Let $p_i = x_{2i+1} - x_{2i-1}$. Note that $$\sum_{i=1}^n p_i = 1.$$ Denote $a_i = x_{2i} - x_{2i-1}$, $b_i = x_{2i+1} - x_{2i}$. We have
$$ x_{2i} = \sum_{j < i} p_j + a_i. $$
Now $a_i + b_i = p_i$ and $0 \leq a_i,b_i \leq h$. Hence
$$ |a_i - p_i/2| = |(b_i-a_i)/2| \leq h/2. $$
Define $a_i = p_i/2 + \delta_i$, so $|\delta_i| \leq h/2$ and
$$ x_{2i} = \sum_{j < i} p_j + p_i/2 + \delta_i. $$
Now we're in good shape since
$$
\begin{align*}
\sum_{i=1}^n x_{2i} (x_{2i+1} - x_{2i-1}) &=
\sum_{i=1}^n \sum_{j<i} p_i p_j + \frac{1}{2} \sum_{i=1}^n p_i^2 + \sum_{i=1}^n \delta_i p_i \\ &=
\frac{1}{2}\left(\sum_{i=1}^n p_i\right)^2 + \sum_{i=1}^n \delta_i p_i \\ &=
\frac{1}{2} + \sum_{i=1}^n \delta_i p_i.
\end{align*}
$$
Since $\sum_{i=1}^n p_i = 1$ and $|\delta_i| \leq h/2$, we deduce
$$
\left|\sum_{i=1}^n x_{2i} (x_{2i+1} - x_{2i-1}) - \frac{1}{2}\right| \leq \frac{h}{2}.
$$
