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

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

• Welcome to MSE! Better use \cdot instead of $*$. Also, the fact to be used is quite unclear to me. Oct 10, 2020 at 13:48
• 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 $||x||^2 =1$. Oct 10, 2020 at 13:54
• @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/… Oct 10, 2020 at 13:59
• thanks, the proof in the link is helpful! Oct 10, 2020 at 14:08
• @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. Oct 10, 2020 at 15:03

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:

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

Therefore:

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

Or:

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

Clearly:

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

Therefore:

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

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.

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