I have shown that the function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ with $\displaystyle{f(x):=\sum_{i=1}^nx_i}$ subject to $\displaystyle{\sum_{i=1}^nx_i^2=1}$ has at $\left (-\frac{\sqrt{n}}{2}, -\frac{\sqrt{n}}{2}, \ldots , -\frac{\sqrt{n}}{2}, -\frac{\sqrt{n}}{2}\right)$ a minimum and at the point $\left (\frac{\sqrt{n}}{2}, \frac{\sqrt{n}}{2}, \ldots , \frac{\sqrt{n}}{2}, \frac{\sqrt{n}}{2}\right)$ there is a maximum.
How can we verify that result using the Cauchy-Schwarz inequality?
By Cauchy-Schwarz inequality, $$\left|\sum_{i=1}^n x_i\right|^2=\left|\sum_{i=1}^n (x_i \times 1)\right|^2\leq \left(\sum_{i=1}^n x_i^2\right)\left(\sum_{i=1}^n 1^2\right)=\left(\sum_{i=1}^n x_i^2\right)\times n=n$$ with equality holds if and only if $$(x_1,\ldots,x_n)=k(1,\ldots,1)$$ for some $k\in\mathbb{R}.$ Given the restriction, you can find the $k$.
Remark: I don't think your maximum and minimum points satisfy the constraint.