Question

show\[||x||_1\leq \sqrt{n}||x||_2 \text{ in } \mathbb R^n\]

Answer 1

\[||x||_1=\sqrt{x_1^2+x_2^2+\cdot \cdot \cdot x_n^2}\] ?
what does \[||x||_2=?\]

Answer 2

\[||x||_2=\sqrt{x_1^2+x_2^2+\cdot \cdot \cdot x_n^2}\]
\[||x||_1=\sum|x_k|\]

Answer 3

ok so
\[||x||_2^2=x_1^2+x_2^2+\cdot \cdot \cdot +x_n^2\]

\[||x||_1=\sqrt{x_1^2}+\sqrt{x_2^2}+ \cdot \cdot \cdot \sqrt{x_n^2}\]
i'm still thinking...

Answer 4

\[||x||_1^2=(\sqrt{x^2_1}+\sqrt{x^2_2}+ \cdot \cdot \cdot +\sqrt{x^n_2})^2\]

Answer 5

\[||x||_1^2=\sum|x_k|^2+2\sum_{i\neq j}|xi||xj|=||x||_2^2+2\sum_{i\neq j}|xi||xj|\]

Answer 6

say the
\[\max(x_1,x_2,x_3,...,x_n)=x_i\]
so couldn't we say
\[||x||^2_2<nx_i^2\]
\[||x||^2_1<(n \sqrt{x_i^2})^2\]
but \[n x^2 _i < (n \sqrt{x^2_i})^2\]
so
\[\sqrt{n} x_i< n \sqrt{x^2_i}\]

.....
thinking

Answer 7

i think you can use the max of the vector in R_n somehow to show this.