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Let $n \geq 3$ be an integer and let $x_1, x_2, \ldots, x_n$ be non-negative real numbers such that $x_1=0, x_n=1$. Prove that there exists $j \in\{1,2, \ldots, n-1\}$ for which:
$\left|x_{j+1}+x_{j-1}-2 x_j\right| \geq \frac{4}{n^2} .$
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Define $x_0=0, x_{n+1}=1, y_k=x_{k+1}-x_k$ for $k=0,1,2, \ldots, n$. Then $y_0=y_n=0$. Assume the contrary, i.e. that $\left|x_{j+1}+x_{j-1}-2 x_j\right|=\left|y_j-y_{j-1}\right|<\frac{4}{n^2}$ for each $j \in\{1,2, \ldots, n\}$. Adding the inequalities $y_j-y_{j-1}<\frac{4}{n^2}$ for $j=1,2, \ldots, k$ we get $y_k<\frac{4 k}{n^2}$. Similarly, by adding $y_j-y_{j+1}<\frac{4}{n^2}$ for $j=k, k+1, \ldots, n-1$ we get $y_k<\frac{4(n-k)}{n^2}$. Consider the following two cases:

Case 1. If $n$ odd then
\begin{aligned} 1 &=y_0+y_1+\cdots+y_n=\left(y_1+\cdots+y_{\frac{n-1}{2}}\right)+\left(y_{\frac{n+1}{2}}+\cdots+y_n\right) \\ &<\frac{4}{n^2} \cdot \frac{1}{2} \cdot\left(\frac{n-1}{2} \cdot \frac{n+1}{2} \cdot 2\right)=\frac{n^2-1}{n^2} \end{aligned}

Case 2. If $n$ is even, we get a contradiction as well since
\begin{aligned} 1 &=y_0+y_1+\cdots+y_n=\left(y_1+\cdots+y_{\frac{n}{2}-1}\right)+\left(y_{\frac{n}{2}}+\cdots+y_{n-1}\right) \\ &<\frac{4}{n^2} \cdot \frac{1}{2} \cdot\left(\left(\frac{n}{2}-1\right) \cdot \frac{n}{2}+\frac{n}{2} \cdot\left(\frac{n}{2}+1\right)\right)=1 \end{aligned}

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