Prove $\left\|x\right\|_2\leq\sqrt{n}\left\|x\right\|_\infty$

Question

I can prove other inequalities like $\|x\|_{1}\leq \sqrt{n}\|x\|_{2}$ and $\|x\|_{1}\leq {n}\|x\|_{\infty}$ using cauchy-schwarz and holder's inequality, but I don't know how to prove this one $\left\|x\right\|_2\leq\sqrt{n}\left\|x\right\|_\infty$. Can anybody tell me how to prove this one?

Also, how do I prove inequalities like these ones: $\|x\|_2\leq\|x\|_1$,$\|x\|_\infty\leq\|x\|_2$, $\|x\|_\infty\leq\|x\|_1$?

Answer 1

If $x=(x_1,...,x_n)$ , then there is $j_0 \in \{1,2,...,n\}$ such that $|x_j|=||x||_{\infty}.$

Thus $||x||_2 \le (\sum_{k=1}^n x_{j_0}^2)^{1/2}=\sqrt{n}\left\|x\right\|_\infty$.

Answer 2

The result is immediate since $$x_1^2+...+x_n^2\leq n\max_{i=1,...,n}\{x_i^2\}=n\left(\max_{i=1,...,n} |x_i|\right)^2.$$