Why is this true: $\|x\|_1 \le \sqrt n \cdot \|x\|_2$?

Using Cauchy-Schwarz, we get:

$$ \Vert x\Vert_1= \sum\limits_{i=1}^n|x_i|= \sum\limits_{i=1}^n|x_i|\cdot 1\leq \left(\sum\limits_{i=1}^n|x_i|^2\right)^{1/2}\left(\sum\limits_{i=1}^n 1^2\right)^{1/2}= \sqrt{n}\Vert x\Vert_2 $$

Hint: This is Cauchy-Schwarz inequality applied to the vectors $x=(|x_k|)_{1\leqslant k\leqslant n}$ and $y=(y_k)_{1\leqslant k\leqslant n}$ defined by $y_k=1$ for every $k$.

We will expand and then complete squares Square up both sides and expand then you get $$ \sum _{i=1}^{n} \sum _{j=1}^{n} |x_i| |x_j| \leq n\sum _ {i=1}^{n} x_i^2$$

$$\sum _{i=1}^{n} \sum _{j=1}^{n} |x_i| |x_j|= \sum_ {1\leq i<j\leq n}2|x_i||x_j| + \sum_ {i=1}^{n} x_i^2$$

So it suffices to prove $$\sum_ {1\leq i<j\leq n}2|x_i||x_j|\leq (n-1)\sum _ {i=1}^{n} x_i^2$$ But $$\sum_ {1\leq i<j\leq n}(|x_i|-|x_j|)^2 =\sum_ {1\leq i<j\leq n} (x_i^2+x_j^2 -2|x_i||x_j|)=$$ $$\sum_ {1\leq i<j\leq n} (x_i^2+x_j^2)+\sum_ {1\leq i<j\leq n}( -2|x_i||x_j|)\geq 0$$

But $$\sum_ {1\leq i<j\leq n} (x_i^2+x_j^2)=\sum_ {1\leq i<j\leq n}x_i^2+\sum_ {1\leq i<j\leq n}x_j^2=\sum_ {i=1}^{n-1}x_i^2 (n-i)+\sum_ {j=2}^{n} x_j^2(j-1)= \sum_ {i=2}^{n-1}x_i ^2 (n-i)+\sum_ {i=2}^{n-1}x_i^2 (i-1)+(n-1)x_1^2+x_n^2(n-1)=$$ $$=(n-1)\sum_ {i=1}^{n}x_i^2$$ And we are done