How do I prove this set of inequalities using Cauchy-Schwarz? Hello all I am trying to understand inequalities and their real world applications, typically optimization techniques. 
I got into to college as a math major and this fall would be my first semester. The road is going to be difficult, and I am trying to brush up on my formulas from now until August. I recently stumbled upon the Cauchy-Schwarz inequality and wanted to understand how it works. I have little understanding on equality theories, so please excuse my ignorance.  I found this example online:
Prove:
$$\frac 1{\sqrt{n}}\|x\|_2 \le \|x\|_{\infty}\le \|x\|_2 \le \|x\|_1 \le \sqrt{n}\|x\|_2\le n\|x\|_{\infty}$$
 using the Cuchy-Schwartz inequality.
Can someone explain how Cauchy-Schwarz solves this problem?
 A: $\left\| x \right\|_1=\left|\langle (1,1,\ldots,1),\left(\left| x_1\right|,\ldots,|x_n|\right) \rangle\right|\leq \left\|\langle (1,1,\ldots,1)\right\|_2 \left\|\left(\left| x_1\right|,\ldots,|x_n|\right) \rangle\right\|_2=\sqrt{n}\left\|x\right\|_2$
This is the only part that makes use of Cauchy-Schwarz. The rest can be proven from definition of the corresponding norm directly. Give it a try. =)
Additional explanation:
Given 2 vectors, $u$ and $u$, Cauchy-Schwarz promises us that the absolute value of inner product (or dot product of two vector) $\left|\langle u, v \rangle \right|=|u^Tv|$  is less than equal to the product of the 2-norm of the two vectors, $\left\|u \right\|\left\| v \right\|$. 
My goal is to show that $\left\|x \right\|_1 \leq \sqrt{n} \left\|x\right\|_2$.
Now, let's see, what does $\left\|x \right\|_1$ means? $\left\|x \right\|_1=\sum_{i=1}^n \left|x_i \right|$. For example  $\left\|\left(\begin{array}{c}1 \\2 \\3 \end{array}\right) \right\|_1=|1|+|2|+|3|$
But I can also write $\left\|x \right\|_1=(1,\ldots,1)\left(\begin{array}{c} |x_1| \\ \vdots \\|x_3| \end{array}\right)$. In my numerical example, I am just saying that $\left\|\left(\begin{array}{c}1 \\2 \\3 \end{array}\right) \right\|_1= (1,1,1)\left(\begin{array}{c} |1| \\ |2| \\|3| \end{array}\right)$. 
That is I have expressed the 1-norm of $x$ as the inner product of $\left(\begin{array}{c} 1 \\ \vdots \\1 \end{array}\right)$ and $\left(\begin{array}{c} |x_1| \\ \vdots \\|x_3| \end{array}\right)$.  In my example, I have expressed $\left\|\left(\begin{array}{c}1 \\2 \\3 \end{array}\right) \right\|_1$ as inner product of $\left(\begin{array}{c} 1 \\ 1 \\1 \end{array}\right)$.  and $\left(\begin{array}{c} |1| \\ |2| \\|3| \end{array}\right)$. 
The moment I have expressed something as a dot product, I can find an upper bound for this quantity using Cauchy-Schwarz. They are the products of the corresponding 2-norm of the two vectors.
The 2-norm of $\left(\begin{array}{c} 1 \\ \vdots \\1 \end{array}\right)$, $ \left\|\left(\begin{array}{c} 1 \\ \vdots \\1 \end{array}\right) \right\|_2=\sqrt{\sum_{i=1}^n 1^2}=\sqrt{n}$.
Hence Cauchy-Schwarz tells us that $\left\| x\right\|_1 \leq \sqrt{n} \left\| x\right\|_2$.
$\left\|\left(\begin{array}{c}1 \\2 \\3 \end{array}\right) \right\|_1=\left|\left(\begin{array}{c}1 \\1 \\1 \end{array}\right)^T \left(\begin{array}{c}1 \\2 \\3 \end{array}\right) \right| \leq \left\|\left(\begin{array}{c}1 \\1 \\1 \end{array}\right) \right\|_2 \left\| \left(\begin{array}{c}1 \\2 \\3 \end{array}\right) \right|_2=\sqrt{3}\left\| \left(\begin{array}{c}1 \\2 \\3 \end{array}\right) \right|_2$
