How can we prove this special case of $\text{A.M.-G.M.}$ Inequality, that is:
If The Geometric Mean of $n$ positive real numbers is equal to $1.$ Prove that their Arithmetic Mean is greater than or equal to $1.$
I know the proof by induction but I want to know if there is a proof without using induction or calculus. Note:I don't want a proof for general $\text{A.M.-G.M.}$ Inequality, I just want a proof for this special case. Any help would be appreciated.
$$\mu=\frac{x_1+x_2+\cdots+x_n}{n}$$ $$l=\sqrt[n]{x_1\,x_2\,\cdots\,x_n}$$ we know $e^x>1+x$. For $i=1,2,\cdots,n$ set $$x=\frac{x_i}{\mu}-1$$ we have $$\large e^{\frac{x_i}{\mu}-1}\ge \frac{x_i}{\mu} $$ therefore \begin{align} & \prod\limits_{i=1}^{n}{{{e}^{\frac{{{x}_{i}}}{\mu }-1}}}\ge \prod\limits_{i=1}^{n}{\frac{{{x}_{i}}}{\mu }}\,\,\,\,\,\Rightarrow \,\,\,{{e}^{-n+\sum\limits_{i=1}^{n}{\frac{{{x}_{i}}}{\mu }}}}\ge \frac{\prod\limits_{i=1}^{n}{{{x}_{i}}}}{{{\mu }^{n}}} \\ \end{align} and $$1\ge \frac{{{l}^{n}}}{{{\mu }^{n}}}\,\,\,\,\,\Rightarrow \,\,{{\mu }^{n}}\ge {{l}^{n}}\Rightarrow \mu \ge l$$