In the proof that $L^{1}$ norm and $L^{2}$ norm are equivalent. To  prove  that  $L^{1}$  norm , denoted  by  $||\ \ ||_{1}$  and  $L^{2}$  norm , denoted  by  $||\ \ ||_{2}$  are  equivalent we  have  to  find  constants   $C_{1},\ \ C_{2}$   that  satisfies   the  following :$$C_{1} ||x||_{1}\le ||x||_{2}\le C_{2} ||x||_{1}$$
Or  according to  my  book  I  have  to  prove  that  $${{1}\over {\sqrt{n}}}\ \ ||x||_{1} \le\ \  ||x||_{2} \ \ \le \ \sqrt n\ \  ||x||_{1}$$
But  I  am  getting  a  different  result. 
  $$||x||_{1}=\sum_{i=1}^{n} |x_{i}|  \ \ \ ;\ \ \ \ \ \  ||x||_{2}=\sqrt{ \sum_{i=1}^{n} x_{i}^{2}}$$
Then $$||x||_{1}^{2} = \sum _{i=1}^{n} |x_{i}|^2 \ge \sum_{i=1}^{n}{x_{i}^{2}}= ||x||_{2}^{2}$$  or  $$||x||_{1}\ge ||x||_{2}$$ On  the  other  hand , we  know  $$|x_{i}|\le ||x||_{2}$$i.e. $$\sum_{i=1}^{n} {|x_{i}}| \le n ||x||_{2}$$ or  $$||x||_{1} \le  n||x||_{2}$$
 So ,  combining  them  we  get
$$||x||_{1} \ \le \ \  n||x||_{2}\ \ \le\ \  n||x||_{1}$$ or $${{1}\over{ n}}||x||_{1} \ \le \ \ ||x||_{2}\ \ \le\ \ ||x||_{1}$$
So  did  I  make  any  mistake  or  that  was  a  simple  printing  mistake  in  my  book $?$
Thanks  for  your  help.
 A: Inequality $\frac{1}{\sqrt{n}}\|x\|_1 \leq \|x\|_2$ can be proven by Cauchy-Schwarz inequality: 
$$\|x\|_1=\sum_{i=1}^n |x_i|\cdot 1 \leq \sqrt{\sum_{i=1}^n x_i^2}\sqrt{\sum_{i=1}^n 1^2}=\sqrt{n}\|x\|_2.$$
A: First I want to say that your results are not wrong, you even find a better constant for $||x||_2 \le ||x||_1$. However, the constant for $C_1 ||x||_1 \le ||x||_2$ can be made more strict.
For the result in the book:
Note that we have
\begin{eqnarray}
||x||_1^2 &=& \left( \sum_{i=1}^n |x_i| \right)^2 \\
&=& \sum_{i,j=1}^n |x_i | | x_j | \\
&=& \sum_{i=1}^n |x_i|^2 + \sum_{i=1}^n \sum_{j \neg i} |x_i | |x_j| \\
&\le& \sum_{i=1}^n |x_i|^2 + \frac{1}{2} \sum_{i=1}^n \sum_{j \neg i} |x_i|^2 + |x_j|^2 \\
&=& \sum_{i=1}^n |x_i|^2 + (n-1) \sum_{i=1}^n |x_i|^2 \\
&=& n ||x||_2^2
\end{eqnarray}
The first inequality is from $a^2+b^2 \ge 2ab$, and we note that $|x_k|^2$ appears $2(n-1)$ times in $\sum_{i=1}^n \sum_{j \neg i} |x_i|^2 + |x_j|^2 $, namely $n-1$ times when $i=k$ and one time when $i\neq k$, which happens $n-1$ times.
So we have $\frac{1}{\sqrt{n}} ||x||_1 \le ||x||_2$.
For the other direction we have
\begin{eqnarray}
||x||_1^2 &=& \sum_{i=1}^n \sum_{j=1}^n |x_i||x_j| \\
&\ge& \sum_{i=1}^n |x_i|^2 \\
&=& ||x||_2^2
\end{eqnarray}
Where the inequality follows from the fact $|x_i| |x_j| \ge 0$ for all $(i,j)$ such that $i \neq j$, so their sum is also non-negative.
So if you want the $\sqrt{n}$, just note that $\sqrt{n} \ge 1$, so we have $\sqrt{n} ||x||_1 \ge ||x||_1 \ge ||x||_2$.
