# 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

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Competition questions are best solved by oneself! That's the only way to get better in solving them. As they say, practice makes perfect. For this particular question, the trick is to convert the "difference sequence" presentation to "partial summation", which is more useful here. –  Yuval Filmus Feb 18 '12 at 2:47
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 $$\left|\sum_{i=1}^n x_{2i} (x_{2i+1} - x_{2i-1}) - \frac{1}{2}\right| \leq \frac{h}{2}.$$