Proving Inequalities: AM-GM-HM, Cauchy-Schwarz/Rearrangement Inequalities.. Okay so I have two questions (I think they are pretty simple which was why I put them together), both relating to inequalities that are proving to be challenging. I have learned the AM-GM-HM Inequalities, the Rearrangement Inequality and the Cauchy-Schwarz Inequality.
A. Show that for positive reals $a, b, c$, such that $abc \le 1$
$$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$$
I have combed through this website and Google for solutions, hints anything but have for the most part come up empty handed. I have seen solutions very similar to it, but none have helped. I saw this exact solution:

It is easy to prove that
$$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge \frac{a + b + c}{\sqrt[3]{abc}}$$
and since $abc \le 1$then $\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$ as required.

It is not clear to me how they arrived at this in particular $\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge \frac{a + b + c}{\sqrt[3]{abc}}$. I feel like it is obvious, but I cant see it. So if someone could explain what inequality/trick was used, I would be grateful.
My attempt (which I think is completely wrong) is as follows.

$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$
by AM-GM inequality $\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge 3\sqrt[3]{\frac{a}{c}\cdot\frac{b}{a}\cdot\frac{c}{b}} = 3$
also by AM-GM $a+b+c \ge 3\sqrt[3]{abc}$
but $abc\le 1$
$\implies 3\sqrt[3]{abc} \le 3$
and thus
$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \ge a + b + c$

B. Let $a_1, a_2, ... , a_n$ be distinct positive integers. Prove that
$$\frac{a_1}{1^2} + \frac{a_2}{2^2} + ... + \frac{a_n}{n^2} \ge \frac{1}{1} + \frac{1}{2} +...+ \frac{1}{n}$$
A friend of mine decided to use the Rearrangement Inequality, but I don't really see that right off the bat. I tried to use Cauchy/Schwarz but I didn't get too far. If someone could give a hint or a nudge in the right direction as to which inequality I should use for this one I would appreciate it.


Show that for positive reals, $a, b, c$ such that $abc \le 1$
  $$\frac{a}{c} + \frac{c}{b} + \frac{b}{a} \ge a+b+c$$

ANS:
Since $abc \le 1 \implies \frac{1}{abc} \ge 1$ and thus 
$$\frac{1}{ac}\ge b, \space \frac{1}{ab}\ge c \space\text{and} \frac{1}{bc}\ge a$$
Now consider $\frac{a}{c} + \frac{a}{c} + \frac{c}{b}$
Using the AM-GM Inequality this gives
$$\frac{a}{c} + \frac{a}{c} + \frac{c}{b} \ge 3\sqrt[3]{\frac{a}{c} \cdot \frac{a}{c} \cdot \frac{c}{b}} = 3\sqrt[3]{\frac{a^2}{bc}}$$
But from above, $\frac{1}{bc} \ge a$
$$\implies \frac{a}{c} + \frac{a}{c} + \frac{c}{b} \ge 3\sqrt[3]{\frac{a^2}{bc}} \ge 3\sqrt[3]{a^3} = 3a$$
Now also consider $\frac{b}{a} + \frac{b}{a} + \frac{a}{c}$ and $\frac{c}{b} + \frac{c}{b} + \frac{b}{a}$ and apply the AM-GM Inequality and use $\frac{1}{ac}\ge b$ and $\frac{1}{ab}\ge c$ from above, we get
$$\implies \frac{b}{a} + \frac{b}{a} + \frac{a}{c} \ge 3\sqrt[3]{\frac{b^2}{ac}} \ge 3\sqrt[3]{b^3} = 3b$$
$$\implies \frac{c}{b} + \frac{c}{b} + \frac{b}{a} \ge 3\sqrt[3]{\frac{c^2}{ab}} \ge 3\sqrt[3]{c^3} = 3c$$
Adding both sides:
$$\frac{a}{c} + \frac{a}{c} + \frac{c}{b} +\frac{b}{a} + \frac{b}{a} + \frac{a}{c} +\frac{c}{b} + \frac{c}{b} + \frac{b}{a} \ge 3a+3b+3c
\\3\left( \frac{a}{c} +\frac{c}{b}+\frac{b}{a}\right) \ge 3(a+b+c)$$

Let $a_1, a_2, ... , a_n$ be distinct positive integers. Prove that
  $$\frac{a_1}{1^2} + \frac{a_2}{2^2} + ... + \frac{a_n}{n^2} \ge \frac{1}{1} + \frac{1}{2} +...+ \frac{1}{n}$$

ANS:
Let $a_1, a_2,...,a_n$ and $\frac{}{}, \frac{}{},...,\frac{}{}$ be two sequences of reals and $a_1', a_2',...,a_n'$ be a permutation of $a_1, a_2,...,a_n.$ Now since $a_1, a_2,...,a_n$ are distinct positive integers, then $a_i' \ge i \forall i$
Now applying the Rearrangement Inequality we get,
$$\frac{a_1}{1^2} + \frac{a_2}{2^2} + ... + \frac{a_n}{n^2} \ge \frac{a_1'}{1^2} + \frac{a_2'}{2^2} + ... + \frac{a_n'}{n^2} \ge \frac{1}{1^2} + \frac{2}{2^2} + ... + \frac{n}{n^2} \ge \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} $$
