Let $M$ be $\max\{x_1,x_2,x_3,\ldots,x_n\}$ and let $m$ be $\min\{x_1,x_2,x_3,\ldots,x_n\}$ .

Then $$M=\frac{M+M+M+\cdots+M}{n}\geq\frac{x_1+x_2+x_3+\cdots+x_n}{n}$$ $$m=\frac{n}{\frac{n}{m}}= \frac{n}{\frac{1}{m}+\frac{1}{m}+\frac{1}{m}+\cdots+\frac{1}{m}} \leq\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\cdots+\frac{1}{x_n}}$$ where all the summations have $n$ terms. So we have proved in this way the two inequalities at the extremes.

Now we shall prove the inequality between arithmetic mean and geometric mean.

Case $n=2$

\begin{eqnarray*} (\sqrt{x_1}-\sqrt{x_2})^2 &\geq& 0\\ x_1-2\sqrt{x_1x_2}+x_2&\geq&0\\ x_1+x_2&\geq&2\sqrt{x_1x_2}\\ \frac{x_1+x_2}{2}&\geq&\sqrt{x_1x_2} \end{eqnarray*}

Case $n=2^k$

\begin{eqnarray*} \lefteqn{ \frac{x_1+x_2+\cdots+x_{2^k}+x_{2^k+1}+\cdots+x_{2^{k+1}}}{2^{k+1}} }\\ &=& \frac{\left( \frac{x_1+x_2+\cdots+x_{2^k}}{2^k} \right) + \left( \frac{x_{2^k+1}+x_{2^k+2}+\cdots+x_{2^{k+1}}}{2^k} \right)} {2} \end{eqnarray*}and using the case $n=2$ on the last expression we can state the following inequality \begin{eqnarray*} \lefteqn{ \frac{x_1+x_2+\cdots+x_{2^k}+x_{2^k+1}+\cdots+x_{2^{k+1}}}{2^{k+1}} }\\ &\ge& \sqrt{ \left( \frac{x_1+x_2+\cdots+x_{2^k}}{2^k} \right) \left( \frac{x_{2^k+1}+x_{2^k+2}+\cdots+x_{2^{k+1}}}{2^k} \right) }\\ &\ge& \sqrt{ \sqrt[2^k]{x_1x_2\cdots x_{2^k}} \sqrt[2^k]{x_{2^k+1}x_{2^k+2}\cdots x_{2^{k+1}}} } \end{eqnarray*}where the last inequality was obtained by applying the induction hypothesis with $n=2^k$ . Finally, we see that the last expression is equal to $\sqrt[2^{k+1}]{x_1x_2x_3\cdots x_{2^{k+1}}}$ and so we have proved the truth of the inequality when the number of terms is a power of two.

Inequality for $n$ numbers implies inequality for $n-1$

Suppose that $$\frac{x_1+x_2+\cdots+x_n}{n}\geq\sqrt[n]{x_1x_2\cdots x_n}$$ is known for a given value of $n$ (we just proved that it is true for powers of two, as example). Then we can replace $x_n$ with the average of the first $n-1$ numbers. So \begin{eqnarray*} \lefteqn{ \frac{ x_1+x_2+\cdots+x_{n-1}+ \left(\frac{x_1+x_2+\cdots+x_{n-1}}{n-1} \right) }{n} }\\ &=& \frac{(n-1)x_1+(n-1)x_2+\cdots+(n-1)x_{n-1}+x_1+x_2+\cdots+x_{n-1}}{n(n-1)}\\ &=& \frac{n x_1+ n x_2 + \cdots + n x_{n-1}}{n(n-1)}\\ &=& \frac{x_1+x_2 + \cdots + x_{n-1}}{(n-1)} \end{eqnarray*} On the other hand \begin{eqnarray*} \lefteqn{\sqrt[n]{ x_1x_2\cdots x_{n-1} \left(\frac{x_1+x_2+\cdots+x_{n-1}}{n-1}\right) }}\\ &=& \sqrt[n]{x_1x_2\cdots x_{n-1}} \sqrt[n]{\frac{x_1+x_2+\cdots+x_{n-1}}{n-1}} \end{eqnarray*}which, by hypothesis (the inequality holding for $n$ numbers) and the observations made above, leads to: $$\left(\frac{x_{1}+x_{2}+\cdots+x_{n-1}}{n-1}\right)^{n}\ge (x_{1}x_{2}\cdots x_{n})\left(\frac{x_1+x_2+\cdots+x_{n-1}}{n-1}\right) $$ and so $$ \left(\frac{x_{1}+x_{2}+\cdots+x_{n-1}}{n-1}\right)^{n-1}\ge x_{1}x_{2}\cdots x_{n} $$ from where we get that $$ \frac{x_{1}+x_{2}+\cdots+x_{n-1}}{n-1}\ge\sqrt[n-1]{x_{1}x_{2}\cdots x_{n}}.$$

So far we have proved the inequality between the arithmetic mean and the geometric mean. The geometric-harmonic inequality is easier. Let $t_i$ be $1/x_{i}$ .

From $$\frac{t_{1}+t_{2}+\cdots+t_{n}}{n}\geq\sqrt[n]{t_{1}t_{2}t_{3}\cdots t_{n}}$$ we obtain $$\frac{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\cdots+\frac{1}{x_{n}}}{n}\geq \sqrt[n]{\frac{1}{x_{1}}\frac{1}{x_{2}}\frac{1}{x_{3}}\cdots\frac{1}{x_{n}}}$$ and therefore $$\sqrt[n]{x_{1}x_{2}x_{3}\cdots x_{n}}\geq \frac{n}{ \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\cdots+\frac{1}{x_{n}} }$$ and so, our proof is completed.