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Given: $\{x_1,x_2,\ldots,x_n\}\subset \Bbb R^+$ and $$\text{(AM)}\ \ \ \frac{x_1+x_2+\ldots+x_n}{n}=\sqrt[n]{x_1 x_2 \ldots x_n}\ \ \ \text{(GM)}$$

Prove: $x_1=x_2=\ldots=x_n$

Background: I am a 9th grader with some experience in math contests. I always see this result but was wondering on how to prove it using simple arguments in the general $n>2$. I know how to prove the converse... it's straightforward.

My attempt (for $n=2$):

It is given that:

$$\frac{x_1+x_2}{2}=\sqrt[2]{x_1 x_2}$$

Squaring both terms and rearranging leads to:

$$x_1^2+x_2^2+2x_1x_2=4x_1x_2 \ \ \Rightarrow\ \ x_1^2+x_2^2-2x_1x_2=0 \ \ \Rightarrow\ \ (x_1-x_2)^2=0$$

This last result is true only if: $x_1=x_2$, completing the proof.

Question: How to prove when $n>2$ ? (9th grader understandable arguments please)

M. Winter
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karina.kawa
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    Excellent question and nice argument. Keep studying. – Randall Sep 26 '17 at 23:16
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    Idle thought: it can be trivially rewritten as $(x_1+x_2+\ldots x_n)^n=n^nx_1x_2\ldots x_n$. The LHS is the number of ways of selecting $n$ elements from a set of size $(x_1+x_2+\ldots+x_n)$; the RHS is the number of ways of taking one element from a set of size $x_1$, one from a set of size $x_2$, etc. times the number of functions from $n$ to $n$. Can you see a 'combinatorial injection' here? – Steven Stadnicki Sep 26 '17 at 23:18
  • Do you consider a proof by induction sufficiently simple? – stewbasic Sep 26 '17 at 23:18
  • @G.Sassatelli Thanks, my mistake. I've just corrected to GM. – karina.kawa Sep 26 '17 at 23:20
  • @stewbasic I understand induction proofs. – karina.kawa Sep 26 '17 at 23:21
  • There are several proofs here, some of them using nothing more complicated than induction: https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Proofs_of_the_AM.E2.80.93GM_inequality – stewbasic Sep 26 '17 at 23:39
  • If you generalize to allow weights, then the inductive step becomes easier to handle. So, by induction, prove: given $t_1, \ldots, t_n > 0$ and $t_1 + \cdots + \cdots t_n = 1$, then $x_1^{t_1} \cdots x_n^{t_n} \le t_1 x_1 + \cdots + t_n x_n$, with equality if and only if $x_1 = \cdots = x_n$. (Admittedly, the $n=2$ case might require more machinery here, though.) – Daniel Schepler Sep 26 '17 at 23:42
  • I've edited this simple inductive proof of the AM-GM inequality to include a proof of the equality condition. – bof Sep 27 '17 at 00:21
  • Another possible approach: first prove for the case when $n$ is a power of two. Then if $n$ isn't a power of two, let $\gamma$ be the geometric mean, and apply the AM-GM inequality (and equality condition) to $x_1, \ldots, x_n, \gamma, \ldots, \gamma$ with $\gamma$ repeated $2^k - n$ times. – Daniel Schepler Sep 27 '17 at 00:45
  • For better search results please add Arithmetic Mean and Geometric Mean in your question (Not only AM and GM). – Royi Sep 27 '17 at 12:29
  • @StevenStadnicki, interesting idea. But even if you get a combinatorial injection, isn't there a lot of extra work to show things extend from positive integers to positive reals? – Barry Cipra Sep 27 '17 at 12:51
  • @BarryCipra No, I think. A proof for positive integers would immediately extend to positive rational numbers, and then to real numbers by rational approximation. – Cave Johnson Sep 27 '17 at 13:05

3 Answers3

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This answer presumes $\text{AM}\ge\text{GM}$ is true. In other words, we consider $\text{AM}\ge\text{GM}$ as a black box, which we uses without proof. If we already know some proof of this inequality, however, the result would immediately follow from it.
Now suppose for some $x_1\ne x_2$ we have $\text{AM}(x_1,x_2,\cdots,x_n)=\text{GM}(x_1,x_2,\cdots,x_n)$. Let $x_1'=x_2'=(x_1+x_2)/2$, we see $$\text{AM}(x_1',x_2',\cdots,x_n)=\text{AM}(x_1,x_2,\cdots,x_n)$$ Meanwhile, we have $x_1'x_2'-x_1x_2=(x_1-x_2)^2/4>0$, hence $$\text{GM}(x_1',x_2',\cdots,x_n)>\text{GM}(x_1,x_2,\cdots,x_n)$$ Combining these results, we find $$\text{AM}(x_1',x_2',\cdots,x_n)<\text{GM}(x_1',x_2',\cdots,x_n)$$ Contradicting $\text{AM}-\text{GM}$ inequality. So $x_1=x_2$ if the equality holds. For the same reason, all $x_i$ is equal.

Cave Johnson
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2

This is not a proof for the general $n$ case, but I think it is a useful result for you to know, using methods you will most surely understand with a little thinking.

Proof: $(n=3)$

As for the 3 variable (all positive) case you can prove using a well know (and useful) identity:

$$a^3+b^3+c^3 - 3abc=\frac{1}{2}(a+b+c)[(a-b)^2+(a-c)^2+(b-c)^2]\ \ \ (1)$$

Make a change of variables in (1), letting $a=\sqrt[3]{x_1}$ $b=\sqrt[3]{x_2}$ and $c=\sqrt[3]{x_3}$, what leads to

$x_1+x_2+x_3 - 3\sqrt[3]{x_1 x_2 x_3}=\frac{1}{2}(\sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_2})[(\sqrt[3]{x_1}-\sqrt[3]{x_2})^2+(\sqrt[3]{x_1}-\sqrt[3]{x_3})^2+(\sqrt[3]{x_2}-\sqrt[3]{x_3})^2]\ \ \ (2)$

We assumed that AM=GM, that is $\frac{x_1+x_2+x_3}{3}=\sqrt[3]{x_1 x_2 x_3}$, and this allows us to conclude that the term on the left hand side (LHS) must be equal to zero.

As $x_1,x_2,x_3$ are all positive, for the term in the right hand side (RHS) be zero, it must be true that

$$(\sqrt[3]{x_1}-\sqrt[3]{x_2})^2+(\sqrt[3]{x_1}-\sqrt[3]{x_3})^2+(\sqrt[3]{x_2}-\sqrt[3]{x_3})^2=0$$

and that will only happen when $x_1=x_2=x_3$. Done!

bluemaster
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Here is a hint for the proof of $n=4$

$(x_1x_2x_3x_4)^{1/4} = \sqrt{\sqrt{x_1x_2}\sqrt{x_3x_4}}$

Can you extend that to $n=8$ and so on?

Then, if you have it for $n=4$, you can show it for $n=3$ by letting $x_4$ be the average of the other three.

Empy2
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