Show that $\max_{1\leq i\leq n}|X_i-\bar{X}| < \frac{(n-1)S}{\sqrt{n}}$ 
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
$$
|X_j-\bar{X}|\le \sqrt{\frac{n-1}{n}}\left(\sum_{i=1}^n (X_i-\bar{X})^2\right)^{1/2}=\frac{(n-1)S}{\sqrt{n}}.
$$ 
