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How can we prove this special case of 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 using only The Order and Field axioms for real numbers or Maybe Least Upper bound axiom. Note:I don't want a proof for general AM-GM Inequality, I just want a proof for this special case. Any help would be appreciated.

420
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3 Answers3

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So, this ends up being a proof of the full AM-GM. Oh well.

Suppose first that $n$ is a power of $2$, and let $n = 2^m$. The term $x_1^n+\dots+x_n^n - nx_1x_2\dots x_n$ can be written as a sum of squares as follows: $$x_1^n+\dots+x_n^n - nx_1x_2\dots x_n = \sum\limits_{k=0}^{m-1}{2^k\sum\limits_{j=0}^{2^{m-k-1}-1}{\left(\prod\limits_{i=1}^{2^k}{x_{2^k(2j)+i}^{2^{m-k-1}}}-\prod\limits_{i=1}^{2^k}{x_{2^k(2j+1)+i}^{2^{m-k-1}}}\right)^2}}$$ or, written less formally, \begin{align} x_1^n+\dots+x_n^n - nx_1x_2\dots x_n =& (x_1^{n/2}-x_2^{n/2})^2+(x_3^{n/2}-x_4^{n/2})^2 + \dots \\ &+2[(x_1^{n/4}x_2^{n/4}-x_3^{n/4}x_4^{n/4})^2 + (x_5^{n/4}x_6^{n/4}-x_7^{n/4}x_8^{n/4})^2+\dots] \\ &+\dots \\ &+\frac{n}{4}[(x_1^2x_2^2\dots x_{n/4}^2 - x_{n/4+1}^2\dots x_{n/2}^2)^2 + (x_{n/2+1}^2\dots x_{3n/4}^2 - x_{3n/4+1}^2+\dots x_n^2)^2] \\ &+\frac{n}{2}(x_1x_2\dots x_{n/2} - x_{n/2+1}\dots x_n)^2. \end{align} For example, if $n=4\implies m=2$, then \begin{align} x_1^4+x_2^4+x_3^4+x_4^4-4x_1x_2x_3x_4 &= \sum\limits_{k=0}^{1}{2^k\sum\limits_{j=0}^{2^{1-k}-1}{\left(\prod\limits_{i=1}^{2^k}{x_{2^k(2j)+i}^{2^{1-k}}}-\prod\limits_{i=1}^{2^k}{x_{2^k(2j+1)+i}^{2^{1-k}}}\right)^2}} \\ &=1\left(\sum\limits_{j=0}^{1}{\left(\prod\limits_{i=1}^{1}{x_{2j+i}^2}-\prod\limits_{i=1}^{1}{x_{2j+1+i}^2}\right)^2}\right) \\ &+ 2\left(\sum\limits_{j=0}^{0}{\left(\prod\limits_{i=1}^{2}{x_{2(2j)+i}} - \prod\limits_{i=1}^{2}{x_{2(2j+1)+i}}\right)^2}\right) \\ &=\left[(x_1^2-x_2^2)^2+(x_3^2-x_4^2)^2\right] + 2(x_1x_2-x_3x_4)^2. \end{align} One can check that for $n=8$ we obtain \begin{align*} x_1^8+x_2^8+\dots+x_8^8 - 8x_1x_2\dots x_8 =& (x_1^4-x_2^4)^2+(x_3^4-x_4^4)^2+(x_5^4-x_6^4)^2+(x_7^4-x_8^4)^2 \\ &+2(x_1^2x_2^2-x_3^2x_4^2)^2+2(x_5^2x_6^2-x_7^2x_8^2)^2 \\ &+4(x_1x_2x_3x_4-x_5x_6x_7x_8)^2. \end{align*} The idea is the following: we try to "complete the square" on terms like $x_1^n+x_2^n$ to yield $x_1^n+x_2^n = (x_1^{n/2}-x_2^{n/2})^2 + 2x_1^{n/2}x_2^{n/2}$, and similarly for the other terms. After collecting the squares and factoring out a $2$, we are left with $x_1^{n/2}x_2^{n/2} + x_3^{n/2}x_4^{n/2} + \dots$, so we try the same process, yielding $x_1^{n/2}x_2^{n/2} + x_3^{n/2}x_4^{n/2} = (x_1^{n/4}x_2^{n/4} - x_3^{n/4}x_4^{n/4})^2 + 2x_1^{n/4}x_2^{n/4}x_3^{n/4}x_4^{n/4}$, and so on. We continue this until we are left with a leftover term where all of the exponents are $1$, and this term will precisely be $nx_1x_2\dots x_n$. This process can be made into a rigorous proof by applying the i-word that the OP seems intent on not using.

So with this decomposition as a sum of squares, it is clear that $x_1^n+x_2^n+\dots x_n^n - nx_1x_2\dots x_n\ge 0$, and hence $$\frac{x_1^n+x_2^n+\dots x_n^n}{n}\ge x_1x_2\dots x_n.$$ The inequality follows by replacing $x_i$ with $x_i^{1/n}$.

Now, if $n$ is not a power of $2$, let $n'$ be a power of $2$ greater than $n$. If $s = \frac{x_1+\dots+x_n}{n}$, then if we consider the sequence $\{x_1,\dots,x_n,s,\dots,s\}$ where $s$ is repeated $n'-n$ times, by AM-GM in the special case we have $$(x_1\dots x_n s\dots s)^{1/n'}\le \frac{x_1+\dots+x_n+s+\dots+s}{n'} = \frac{ns+(n'-n)s}{n'} = s$$ and hence $$x_1\dots x_n s^{n'-n}\le s^{n'}\implies x_1\dots x_n\le s^n$$ thus yielding the full AM-GM inequality.

Joey Zou
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Hint: Equivalently you want to show for positives, $$\prod_k x_k =1 \implies \sum_k (x_k-1) \geqslant 0$$

This would follow if you can show $x-1 \geqslant \log x$ for all positive $x$, which is obvious as $\log x$ is concave so it stays below its tangent $x-1$.

Macavity
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  • How can we prove $x-1>=log x$ algebraically. – 420 Sep 07 '16 at 17:18
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    @user348631 without concavity or calculus / MVT, (I assume you are ruling all of that out by "algebraic") it could be really tiresome. Why is it you want to avoid induction, calculus, concavity etc.? Best put down your complete list of rules and why you think that's sensible. Log definition itself will need induction or limit of series or calculus. I could show a smoothing argument, but you may rule that out too, so will await your allowable set of rules before doing anything more here. – Macavity Sep 07 '16 at 17:28
  • @user348631 your comment question is equivalent to http://math.stackexchange.com/questions/504663/simplest-or-nicest-proof-that-1x-le-ex/504671#504671. Take your pick. – Macavity Sep 07 '16 at 17:34
  • Because all the proof I have seen mostly use induction or calculus.So I wanted to know a self contained algebraic proof. – 420 Sep 07 '16 at 17:36
  • I have updated my question please show a smoothing argument. – 420 Sep 07 '16 at 18:02
  • I have updated my question please show a smoothing argument. – 420 Sep 07 '16 at 18:04
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Let $A=\dfrac{x_1+x_2+\dots+x_n}n$, $\;G=(x_1x_2\dotsm x_n)^{\tfrac1n}$. Since $\log$ is concave, $$\log A=\log\dfrac{x_1+x_2+\dots+x_n}n\le \frac 1n\log x_1+\frac1n\log x_2+\dots+\frac1n\log x_n=\log G$$ by Jensen's inequality.

Of course, there's induction in the proof of Jensen's inequality.

Bernard
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