Prove $(x_1x_2\cdots x_n)^{\frac{1}{n}} \leq \frac{1}{n}(x_1+x_2+\cdots+x_n)$

for all $x_1,\ldots, x_n > 0$.

To prove this we are supposed to use the fact that the maximum of $(x_1x_2\cdots x_n)^2$, for all $x$ with $||x||^2 =1$,

is achieved in the point $a= (\frac{1}{\sqrt{n}}, ... , \frac{1}{\sqrt{n}})$, with the maximum equal to $\frac{1}{n^n}$. The value of point $a$ I calculated using the theorem of lagrange multipliers, however I do not know how to prove the statement with which I started the question.

Big thanks

  • $\begingroup$ Welcome to MSE! Better use \cdot instead of $*$. Also, the fact to be used is quite unclear to me. $\endgroup$
    – VIVID
    Oct 10, 2020 at 13:48
  • $\begingroup$ thanks for the tip! I think what we are supposed to use is that $(x_1 \cdot x_2 \cdot ... \cdot x_n)^2 \leq n$ for all $x$ with $||x||^2 =1$. $\endgroup$
    – Jonas
    Oct 10, 2020 at 13:54
  • $\begingroup$ @Jonas: It’s incorrect when $n\neq 2$. you may want to take a look at this Wikipedia entry regarding this en.m.wikipedia.org/wiki/… $\endgroup$ Oct 10, 2020 at 13:59
  • $\begingroup$ thanks, the proof in the link is helpful! $\endgroup$
    – Jonas
    Oct 10, 2020 at 14:08
  • 1
    $\begingroup$ @Jonas this is the famous am-gm inequality you can search on the site as proof of am -gm inequality I am sure it is present. $\endgroup$ Oct 10, 2020 at 15:03

2 Answers 2


This inequality is known as the inequality of arithmetic and geometric means, or the AM-GM inequality for short.

There are many many ways to prove it and I think you can easily find such proofs on the internet (such as on this website). Though it seems to me that you want a proof that uses the following fact that you are able to prove yourself:

The maximum of $(y_1\dots y_n)^2$ subject to $\|y\|^2=1$ is $1/n^n$, achieved when $y_1=\dots=y_n=1/\sqrt{n}$.

Indeed, applying the above result for $$y_i = \sqrt{\frac{x_i}{x_1+\dots + x_n}}, \quad i=1,2,\dots,n,$$ (notice that $\|y\|^2=1$), we see that the maximum of $\frac{x_1\dots x_n}{(x_1+\dots + x_n)^n}$ is $1/n^n$, which means $$\frac{x_1\dots x_n}{(x_1+\dots + x_n)^n} \le \frac{1}{n^n},$$ or equivalently $$(x_1\dots x_n)^{1/n} \le \frac{x_1+\dots + x_n}{n},$$ achieved when $\sqrt{\frac{x_i}{x_1+\dots + x_n}} = \frac{1}{\sqrt{n}} \ \forall i$, i.e., $x_1=x_2=\dots=x_n$.

P/s: The technique for reducing the original unconstrained inequality over $x$ to the above constrained inequality over $y$ is called normalization.


We know that:

$$(x_1+x_2+x_3+ \cdot \cdot \cdot + x_n)^n= n!(x_1x_2x_3 \cdot \cdot\cdot x_n) + M$$


$$(x_1+x_2+x_3+ \cdot \cdot \cdot + x_n)^n> n!(x_1x_2x_3 \cdot \cdot\cdot x_n)> n(x_1x_2x_3 \cdot \cdot\cdot x_n)$$


$$\frac{1}{(n)^{\frac 1n}}(x_1+x_2+x_3+ \cdot \cdot \cdot + x_n)> (x_1x_2x_3 \cdot \cdot\cdot x_n)^{\frac 1n}$$


$\frac{1}{(n)^{\frac 1n}}>\frac1n$


$$\frac{1}n(x_1+x_2+x_3+ \cdot \cdot \cdot + x_n)\geqslant (x_1x_2x_3 \cdot \cdot\cdot x_n)^{\frac 1n}$$

Equality holds if $x_1=x_2=x_3= \cdot\cdot\cdot x_n$

There is also a proof in:Classical algebra, author: G.Paria, ISBN 81-7381-16-1, page 160.

  • $\begingroup$ Why does $\frac{1}{(n!)^{\frac 1n}}>\frac1n$ imply the last inequality? $\endgroup$
    – f10w
    Oct 10, 2020 at 20:47
  • $\begingroup$ @Jonas, did you find the solution in reference I gave? $\endgroup$
    – sirous
    Oct 11, 2020 at 8:20
  • $\begingroup$ @Khue, I corrected my answer. $\endgroup$
    – sirous
    Oct 11, 2020 at 12:11
  • $\begingroup$ Well, it's even worse. $\frac{1}{(n)^{\frac 1n}}<\frac1n$ is clearly wrong. $\endgroup$
    – f10w
    Oct 11, 2020 at 12:25

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