Problem 1695

For $n \geqslant 2$, let $\sigma(1), \sigma(2), \cdots, \sigma(n)$ be a permutation of $1,2, \cdots, n$, and satisfy $0 \leqslant a_{\sigma(1)}<a_{\sigma(2)}<\cdots<a_{\sigma(n)} \leqslant c$, then
$$\begin{array}{l} c \geqslant a_{\sigma(n)}-a_{\sigma(1)} \geqslant\left(a_{\sigma(n)}-a_{\sigma(n-1)}\right)+ \\ \left(a_{\sigma(n-1)}-a_{\sigma(n-2)}\right)+\cdots+\left(a_{\sigma(2)}-a_{\sigma(1)}\right) \geqslant \\ \frac{1}{\sigma(n)+\sigma(n-1)}+\frac{1}{\sigma(n-1)+\sigma(n-2)}+\cdots+\frac{1}{\sigma(2)+\sigma(1)} \end{array}$$

By the Cauchy-Schwarz inequality
$$\begin{array}{l} {\left[\frac{1}{\sigma(n)+\sigma(n-1)}+\frac{1}{\sigma(n-1)+\sigma(n-2)}+\cdots+\frac{1}{\sigma(2)+\sigma(1)}\right]} \\ {[(\sigma(n)+\sigma(n-1))+(\sigma(n-1)+\sigma(n-2))+\cdots+(\sigma(2)+\sigma(1))] \geqslant} \\ (n-1)^{2} \end{array}$$

we get
$$\begin{aligned} c \geqslant & \frac{(n-1)^{2}}{2[\sigma(1)+\sigma(2)+\cdots+\sigma(n-1)+\sigma(n)]-\sigma(1)-\sigma(n)}= \\ & \frac{(n-1)^{2}}{n(n+1)-\sigma(1)-\sigma(n)} \geqslant \\ & \frac{(n-1)^{2}}{n(n+1)-3} \geqslant \frac{n-1}{n+3}=1-\frac{4}{n+3} \end{aligned}$$

for all positive integers $n \geqslant 2$, hence it must be that $c \geqslant 1$.