Two Part Problem: Prove that $||x||_\infty \leq ||x|| \leq ||x||_1.$ Show from the Schwarz Inequality that the ratios $||x|| / ||x||_\infty$ and $||x||_1 / ||x||$ are never larger than $\sqrt n$. Which vector gives ratios equal to $\sqrt n$?
My Try: Proving the inequality first.
$||x||_\infty \leq ||x|| \leq ||x||_1 = max|x_i| \leq \sqrt{x \cdot x} \leq \sum_{i=1}^{n}|x_i|$
Case 1: $max|x_i| \leq \sum_{i=1}^{n}|x_i|$ \
Because on the left side we're taking 1 element of a set (absolute value). On the right side we're taking the sum of all the elements of the set (absolute value).
Case 2: $\sqrt{x \cdot x} = \sqrt{x_{1}^{2} + .... + x_{n}^{2}}$
$\sqrt{\sum_{i=1}^{n}x_i^2} \leq \sum_{i=1}^{n}|x_i|$
$\sum_{i=1}^{n}x_i^2 \leq (\sum_{i=1}^{n}|x_i|)^2$
Since the sum of squares $\leq$ the square of the sum, this holds. \
Case 3: $max|x_i| \leq \sqrt{x \cdot x}$
$max(|x_i|)^2 \leq x_{1}^{2} + .... + x_{n}^{2}$
This holds because the left side is only 1 element of the set squared, the right side is the sum of all squared elements.
Stuck here
I believe my prove so far is correct. The problem I face is in the 2nd part, where I'm not sure how to use this to prove the ratios are never larger than $\sqrt n$.
Your analysis is correct. As a side note, you only needed to prove cases 2 and 3, as case 1 is a consequence of 2 and 3.
One form of the Cauchy-Schwarz inequality is as follows, \begin{gather} \left(\sum_{i=1}^n a_i^2\right) \left(\sum_{i=1}^n b_i^2\right) \geq \left(\sum_{i=1}^n a_i b_i\right)^2 \end{gather}
Looking at the structure, the first term resembles $||x||$ and the third resembles $||x||_1$. Choose $a_i$ and $b_i$ so that you get $||x||$ and $||x||_1$ in the above.
For $\frac{||x||}{||x||_{\infty}}$, you could prove it by starting from $x_i^2 \leq \max(|x_j|)^2$, and summing over $i$ on both sides. Though I don't know how to use the Schwarz inequality for it.