Define $||x||_p=(|x_1|^p+\cdots+|x_n|^p)^{\frac{1}{p}}$. So, $||\cdot||_2$ is the square root of sum of squares, and $||\cdot||_1$ is the sum of absolute values.
I want to show that
$||x||_2\leq ||x||_1\leq \sqrt{n}||x||_2$.
I proved the first inequality by induction. The second inequality gets me in trouble. Any hint will be appreciated.
Also, what's the geometric intuition of this problem? I guess there's a short cut from a geometric view.
On
Hints:
For the first inequality (even though you have an induction proof), observe that $$ \|x\|_2^2=|x_1|^2+|x_2|^2+\dots+|x_n|^2 $$ while $$ \|x\|_1^2=\left(|x_1|+|x_2|+\dots+|x_n|\right)^2. $$ Then $\|x\|_1^2$ contains all of the terms of $\|x_2\|^2$ and some extra ones (and everything is positive).
For the second inequality, use the Cauchy-Schwarz inequality. In particular $$ \|x\|_1^2=\left|\sum 1\cdot |x_i|\right|^2\leq\left(\sum 1^2\right)\left(\sum |x_i|^2\right)=n\|x\|_2^2. $$
For geometric intuition, I suggest plotting the unit $1$-norm spheres in $2$ dimensions (diamond) and $3$ dimensions (octahedron), then plotting the largest $2$-norm spheres inscribed in them, respectively in $2$ (circle) and $3$ (sphere) dimensions.
Look at where the inscribed $2$-norm spheres touches the $1$-norm spheres: at the points $(\pm 1/2, \pm 1/2)$ and $(\pm 1/3, \pm 1/3, \pm 1/3)$. At these points, the $1$-norm is $1$, and the $2$-norm is $n^{-1/2}$. Therefore, the rest of the $2$-sphere is of radius $n^{-1/2}$. Because this sphere is completely contained in the $1$-ball of radius $1$, we can deduce that $$\|x\|_2 \le n^{-1/2} \implies \|x\|_1 \le 1,$$ for $n = 2, 3$. Therefore, given any $x$, we can consider $y = \frac{x}{\|x\|_2 \sqrt{n}}$, and get $$\|y\|_2 = \left\|\frac{x}{\sqrt{n}\|x\|_2}\right\|_2 = n^{-1/2} \implies \|y\|_1 \le 1 \implies \|x\|_1 \le \sqrt{n}\|x\|_2.$$ Again, this only works for $n = 2$ and $n = 3$. Everything generalises pretty simply to higher dimensions, but it's harder to call it geometric intuition when you can't properly visualise it.