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.


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$$\mu=\frac{x_1+x_2+\cdots+x_n}{n}$$ $$l=\sqrt[n]{x_1\,x_2\,\cdots\,x_n}$$ 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 $$1\ge \frac{{{l}^{n}}}{{{\mu }^{n}}}\,\,\,\,\,\Rightarrow \,\,{{\mu }^{n}}\ge {{l}^{n}}\Rightarrow \mu \ge l$$

  • $\begingroup$ Then how to prove that $e^x>1+x$ if $x>0$. $\endgroup$ – Shubhashish Jun 22 '16 at 12:51
  • 1
    $\begingroup$ $$e^x=1+x+\frac{1}{2!}x^2+...$$ $\endgroup$ – Behrouz Maleki Jun 22 '16 at 12:52
  • $\begingroup$ Small nit, your usage of $e^x$ is actually over all reals not just non-negative, so while $e^x\ge x+1$ for all reals, that proof isn't quite so obvious. $\endgroup$ – Macavity Jun 22 '16 at 16:50
  • $\begingroup$ Why does my solution isn't quite so obvious? $\endgroup$ – Behrouz Maleki Jun 22 '16 at 16:52
  • $\begingroup$ For $x =-0.9$, how does your one liner show $e^x\ge x+1$ ? And you have not even included negative numbers in your statement to start with. $\endgroup$ – Macavity Jun 22 '16 at 16:55

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