Proving HM-GM inequality using AM-GM

Question

I want to prove

$$\frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}} ≤ \sqrt[n]{a_1\cdots a_n}$$

Using AM-GM inequality.

My attempt:

Consider arbitrary $x_1,\cdots,x_n$ We know that

$$\sqrt[n]{x_1\cdots x_n} ≤ \frac{x_1 + \cdots + x_n}{n}$$

Replacing all $x$ with $\frac{1}{x}$

$$\begin{align}\sqrt[n]{\frac{1}{x_1}\cdots \frac{1}{x_n}} ≤ \frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n} & \implies \frac{1}{x_1}\cdots \frac{1}{x_n} ≤ \biggl(\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\biggr)^{n} \\ & \implies \frac{1}{\frac{1}{x_1}\cdots \frac{1}{x_n}} ≥ \frac{1}{\biggl(\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\biggr)^{n}} \\ & \implies x_1\cdots x_n ≥ \frac{1}{\biggl(\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\biggr)^{n}} \\ & \implies \sqrt[n]{x_1\cdots x_n} ≥ \frac{1}{\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}} \\ & \implies \sqrt[n]{x_1\cdots x_n} ≥ \frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}} \end {align}$$

$\Box$

Is it correct? Are there other (better) alternatives?

Answer 1

a)Yes, it is correct.

b) A better method would be to take reciprocals in the first step, as both sides are positive. $$\sqrt[n]{\frac{1}{x_1}\cdots \frac{1}{x_n}} \leq \frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}$$ $$\iff \sqrt[n]{x_1x_2\cdots x_n}\geq\frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}$$

Answer 2

That's pretty much the standard proof that AM-GM implies GM-HM. The only variation among specific presentations of them is how much detail to go into. For example, you could argue you can just write$$\sqrt[n]{\frac{1}{x_1}\cdots \frac{1}{x_n}}\le\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\implies\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}.$$Edit: This is basically the point made in @ViktorGlombik's comment. (Incidentally, you can get $\le,\,\ge$ respectively with \le, \ge.)