Let $X_1, X_2,...,X_n$ be sample from some population. Show that $$\max_{1\leq i\leq n}|X_i-\bar{X}| < \dfrac{(n-1)S}{\sqrt{n}}$$
Simplifying the inequality, I found, to prove: $$n\max_{1\leq i\leq n}(X_i-\bar{X})^2 < (n-1)\sum_{i=1}^n (X_i-\bar{X})^2$$ or more generally, we need to prove, $n\cdot\max_{1\leq i\leq n}Y_i^2 < (n-1) \sum_{i=1}^n Y_i^2$
I am stuck here. Any help appreciated.
For any $t_1,\ldots,t_n$, \begin{align} \left|\sum_{i=1}^n t_i(X_i-\bar{X}) \right|&=\left|\sum_{i=1}^n (t_i-\bar{t})(X_i-\bar{X}) \right| \\ &\le \left(\sum_{i=1}^n (t_i-\bar{t})^2\sum_{i=1}^n (X_i-\bar{X})^2\right)^{1/2}. \end{align} For $j\in \{1,\ldots,n\}$ take $t_j=1$ and $t_k=0$ for $k\ne j$. Then $$ |X_j-\bar{X}|\le \sqrt{\frac{n-1}{n}}\left(\sum_{i=1}^n (X_i-\bar{X})^2\right)^{1/2}=\frac{(n-1)S}{\sqrt{n}}. $$