Question

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:
\[
\left|x_{j+1}+x_{j-1}-2 x_j\right| \geq \frac{4}{n^2} .
\]

Answer 1

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
\[
\begin{aligned}
1 &=y_0+y_1+\cdots+y_n=\left(y_1+\cdots+y_{\frac{n-1}{2}}\right)+\left(y_{\frac{n+1}{2}}+\cdots+y_n\right) \\
&<\frac{4}{n^2} \cdot \frac{1}{2} \cdot\left(\frac{n-1}{2} \cdot \frac{n+1}{2} \cdot 2\right)=\frac{n^2-1}{n^2}
\end{aligned}
\]
a contradiction.

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