Proving HM-GM inequality using AM-GM 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? 

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