 Sharp comparison of l1 and l2 norms

For $x\in {\mathbb{R}}^n$, Cauchy-Schwarz inequality applied to $(|x_i|)$ and $(1)$ gives the upper bound. Equal absolute coordinate values attain its factor $\sqrt{n}$. Expanding $({\sum }_i|x_i|{)}^2$ gives the lower bound, and a vector with only one nonzero coordinate attains its factor one. Thus both constants are optimal. These Lp norms define the same finite-dimensional topology despite having different unit balls.