 Solution

A norm is a function $N:{\mathbb{R}}^n\to [0,\infty )$ such that $N(x)=0$ exactly when $x=0$, $N(tx)=|t|N(x)$, and $N(x+y)\le N(x)+N(y)$. Both specified Lp norms are nonnegative, vanish only at zero and satisfy absolute homogeneity immediately from their formulas.

For the $1$-norm, the scalar triangle inequality gives ${\sum }_i|x_i+y_i|\le {\sum }_i|x_i|+{\sum }_i|y_i|$. For the $2$-norm, Cauchy-Schwarz inequality givesTaking nonnegative square roots proves its triangle inequality. Hence both are norms.

For the sharp comparison of l1 and l2 norms, Cauchy-Schwarz yieldsThe vector $(1,\dots ,1)$ attains equality, so no smaller constant works. Conversely, $({\sum }_i|x_i|{)}^2\ge {\sum }_i|x_i{|}^2$, giving $\Vert x{\Vert }_2\le \Vert x{\Vert }_1$. A standard basis vector attains equality. Thus, for $n\ge 1$,