Is There an Inequality Between L1 and L2 Norms?

[roho]

Homework Statement

$$\|x\|_2\le\|x\|_1\le\sqrt{n}\|x\|_2$$
where |x|1 is the l1 norm and |x|2 is the l2 norm

Homework Equations

See above

The Attempt at a Solution

I have $$\|\mathbf{x}\|_1 := \sum_{i=1}^{n} |x_i|$$
and $$\|x\|_2 = \left(\sum_{i\in\mathbb N}|x_i|^2\right)^{\frac12}$$
I have tried to expand out the x 2 norm but i can't seem to figure out how to prove the inequality. Any suggestions?

 

[lanedance]

for the first part of the inequality, you could try squaring both sides

 

[roho]

Yea that works for the first part. Thanks for the reply.

Any idea on the second part (square root of n)?

I am thinking it may have to do with the projection vector (such as (1,1,1,1,1,1)) in a scalar product or something like.

 

[lanedance]

your idea should work with for the 2nd one with the use of Cauchy Schwarz