<s>[INST] 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.
 [/INST] For the first inequality, the trick is to apply the AM-GM inequality to each of the three terms in $${a\over{\sqrt[3]{abc}}} + {b\over{\sqrt[3]{abc}}}+ {c\over{\sqrt[3]{abc}}} .$$
(mouse over for spoiler)

 $${a\over{\sqrt[3]{abc}}}=\sqrt[3]{a^3\over abc}=\sqrt[3]{\frac ab\cdot\frac ab\cdot\frac bc}$$

Hint for the second inequality: apply the Rearrangement Inequality to $a_1, a_2, \ldots,a_n$ and ${1\over1^2},{1\over2^2},\ldots,{1\over n^2}$. Once that's done, argue that the result follows.
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