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The arithmetic - geometric mean inequality states that $$\frac{x_1+ \ldots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}$$ I'm looking for some original proofs of this inequality. I can find the usual proofs on the internet but I was wondering if someone knew a proof that is unexpected in some way. e.g. can you link the theorem to some famous theorem, can you find a non-trivial geometric proof (I can find some of those), proofs that use theory that doesn't link to this inequality at first sight (e.g. differential equations …)?

Induction, backward induction, use of Jensen inequality, swapping terms, use of Lagrange multiplier, proof using thermodynamics (yeah, I know, it's rather some physical argument that this theorem might be true, not really a proof), convexity, … are some of the proofs I know.

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    $\begingroup$ What have you seen, so that we don't repeat what you already know? $\endgroup$ – chubakueno Feb 26 '14 at 21:45
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    $\begingroup$ Would you include that in your post? (And,thermodynamics? ._. ) $\endgroup$ – chubakueno Feb 26 '14 at 21:54
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    $\begingroup$ I'm curious about the thermodynamics proof $\ddot \smile$ $\endgroup$ – dani_s Feb 26 '14 at 22:08
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    $\begingroup$ Community wiki? $\endgroup$ – abnry Feb 26 '14 at 22:28
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    $\begingroup$ Please add the proofs you know as answers (just to get them all together). $\endgroup$ – vonbrand Feb 27 '14 at 1:25

19 Answers 19

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Pólya's Proof:

Let $f(x) = e^{x-1}-x$. The first derivative is $f'(x)=e^{x-1}-1$ and the second derivative is $f''(x) = e^{x-1}$.

$f$ is convex everywhere because $f''(x) > 0$, and has a minimum at $x=1$. Therefore $x \le e^{x-1}$ for all $x$, and the equation is only equal when $x=1$.

Using this inequality we get

$$\frac{x_1}{a} \frac{x_2}{a} \cdots \frac{x_n}{a} \le e^{\frac{x_1}{a}-1} e^{\frac{x_2}{a}-1} \cdots e^{\frac{x_n}{a}-1}$$

with $a$ being the arithmetic mean. The right side simplifies

$$\exp \left(\frac{x_1}{a} -1 \ +\frac{x_1}{a} -1 \ + \cdots + \frac{x_n}{a} -1 \right)$$

$$=\exp \left(\frac{x_1 + x_2 + \cdots + x_n}{a} - n \right) = \exp(n - n) = e^0 = 1$$

Going back to the first inequality

$$\frac{x_1x_2\cdots x_n}{a^n} \le 1$$

So we end with

$$\sqrt[n]{x_1x_2\cdots x_n} \le a$$

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  • $\begingroup$ Almost all your inequalities are reversed. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 26 '14 at 23:18
  • $\begingroup$ @Martín-BlasPérezPinilla Woops, fixed $\endgroup$ – qwr Feb 26 '14 at 23:23
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    $\begingroup$ This reminds me of a trick I found: $e^x \geq 1+x$ by similar calculations as the start of this answer. Then $e^{1/n} > 1 + 1/n = (n+1)/n,$ so $e^{1 + 1/2 + 1/3 + \ldots + 1/n} > (2/1) (3/2) \ldots ( (n+1)/n ) = n+1,$ so $H_n:= 1 + 1/2 + \ldots + 1/n > \log (n+1),$ and in particular it diverges. $\endgroup$ – Ragib Zaman Jun 12 '14 at 4:17
  • $\begingroup$ @qwr : i know this thread is a few years old but i still have a question. How is exp(xi/a - n)=exp(n-n) ? $\endgroup$ – hukachaka Oct 28 '17 at 14:41
  • $\begingroup$ @hukachaka $a$ is defined as the arithmetic mean. $\endgroup$ – qwr Oct 30 '17 at 7:00
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I shall provide a simple geometric proof of the inequality in the case of two variables (which I have not been able to find anywhere else - a proof involving a triangle in a circle seems to be popular).

Consider the square of side $a + b$ in the figure below.
AM-GM

The area of the square is $(a + b)^2$. But as it completely contains the four blue rectangles, each of area $ab$, it follows that

$(a + b)^2 \ge 4ab \Rightarrow\\ \dfrac{a + b}{2} \ge \sqrt{ab} $

Further, note that there is a square in middle, of side $(b - a)$, and hence area $(b - a)^2$. Therefore the inequality is strict except when $a = b$.

This proves the two-variable case. The same can be extended to the $n$-variable case. I have tried extending it to three variables, but it is difficult to argue why exactly $27$ rectangular parallelepipeds (of sides $a, b, c$) fit in the cube (of side $a + b + c$), though I can see it is so. Any suggestions?

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As requested by dani_s, I will give the thermodynamic proof of the AM-GM inequality. This is certainly an example of an original proof, although you might argue about whether or not it's rigorous.

Let's start with a list of numbers $x_i$ for which we want to prove the inequality. Take $n$ identical heat reservoirs with the same heat capacity $c$. Reservoir $i$ had initial temperature $x_i$. Bring those reservoirs in contact with each other such that this system evolves to an equilibrium temperature A.

The first law of thermodynamics (conservation of energy) implies that A equals the arithmetic mean of the $x_i$, AM.

The second law of thermodynamics states that the entropy increases until the equilibrium is reached, where the entropy has a maximum. The corresponding formula of change in entropy is $$\Delta S=c \ln{\frac{T}{T_0}}$$ where $c$ is the heat capacity, $T_0$ the initial temperature and $T$ the end temperature.

In our case $T_i=A$ for all $i$ and $T_{0,i}=x_i$. The total entropy didn't decrease and therefore, $$\sum_{i=1}^n c \ln\frac{A}{x_i} \geq 0$$

By writing the sum of logarithms as a logarithm of a product, we recognize the geometric mean. Therefore (since $A=AM$): $$\frac{AM^n}{GM^n} \geq 1$$ This proves the AM-GM inequality.

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    $\begingroup$ There is no need to argue whether this is or not a rigorous proof, as it isn't. $\endgroup$ – Mariano Suárez-Álvarez Jun 17 '15 at 16:52
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    $\begingroup$ The assumption is that entropy is minimum when all $T_i$ are equal, is known to be true only because of AGM is known to hold. $\endgroup$ – DanielWainfleet Feb 21 '16 at 4:13
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    $\begingroup$ interesting and fun concept, but nonetheless very circular $\endgroup$ – Charlie Tian Mar 9 '18 at 5:49
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As $(\sqrt{x_1}-\sqrt{x_2})^2 \geq 0$ we have $$\sqrt{x_1 \cdot x_2} \leq \frac{x_1+x_2}{2}.$$ Applying this inequality twice, we get $$(x_1 x_2 x_3 x_4)^{\frac{1}{4}} \leq \frac{\sqrt{x_1 x_2}+\sqrt{x_3 x_4}}{2} \leq \frac{x_1+x_2+x_3+x_4}{4}.$$ By induction, it is not difficult to see that $$(x_1 \cdots x_{2^k})^{\frac{1}{2^k}} \leq \frac{x_1+\ldots+x_{2^k}}{2^k} \tag{1}$$ for all $k \geq 1$.

It remains to fill the gaps between the powers of two. So let $x_1,\ldots,x_n$ be arbitrary positive numbers and choose $k$ such that $n\leq 2^k$. We set

$$\alpha_i := \begin{cases} x_i & i \leq n \\ A & n< i \leq 2^k \end{cases}$$

where $A:= \frac{x_1+\ldots+x_n}{n}$. Applying $(1)$ to the $(\alpha_1,\ldots,\alpha_{2^k})$ yields

$$\bigg( x_1 \ldots x_n A^{2^k-n} \bigg)^{\frac{1}{2^k}} \leq \frac{x_1+\ldots+x_n+(2^k-n) A}{2^k} = A.$$

Hence,

$$(x_1 \ldots x_n)^{1/n} \leq A = \frac{x_1+\ldots+x_n}{n}.$$

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    $\begingroup$ I believe this one is due to Cauchy. $\endgroup$ – Siméon Jun 11 '14 at 19:00
  • $\begingroup$ Applying the general Jensen inequality ($n$ terms, not just two) it gets much simpler, see eg Wikipedia $\endgroup$ – leonbloy Jun 12 '14 at 14:16
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    $\begingroup$ The first inequality is just from $(x_1-x_2)^2 \ge 0$... Why on earth would you use Jensen's which is so much harder to prove? $\endgroup$ – user21820 Jun 8 '16 at 6:02
  • $\begingroup$ Thanks for pointing this out; I changed my answer accordingly. $\endgroup$ – saz Feb 20 '18 at 10:47
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    $\begingroup$ shouldn't the first step be $(\sqrt{x_1} - \sqrt{x_2})^2 \ge 0$? $\endgroup$ – Yingkai Ouyang Jul 30 '18 at 9:06
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LEMMA. If $a_1,\dots,a_n$ are positive numbers whose product is equal to $1,$ then $a_1+\dots+a_n\ge n,$ with equality only when $a_1=\cdots=a_n=1.$

Proof by induction on $n.$ The case $n=1$ being trivial, we suppose $n\ge2.$ Let $a_1,\ a_2,\ \dots,\ a_n$ be positive numbers with $a_1a_2\dots a_n=1.$ Without loss of generality, we can assume that $a_1=\max\{a_1,\dots,a_n\}\ge1$ and $a_2=\min\{a_1,\dots,a_n\}\le1.$ Thus we have $$a_1+a_2-a_1a_2-1=(a_1-1)(1-a_2)\ge0,$$ i.e., $$a_1+a_2-a_1a_2\ge1.\tag1$$ Since $a_1a_2,\ a_3,\ a_4,\ \dots,\ a_n$ are $n-1$ positive numbers with product equal to $1,$ by the inductive hypothesis we have $$a_1a_2+a_3+\cdots+a_n\ge n-1.\tag2$$

Adding the inequalities (1) and (2), we get $$a_1+a_2+a_3+\cdots+a_n\ge1+(n-1)=n.\tag3$$ If equality holds in (3) then we must have equality in (1), that is, $a_1=1$ or $a_2=1,$ whence $a_1=\dots=a_n=1.$


THEOREM. If $x_1,\ x_2,\ \dots,\ x_n$ are positive numbers, then $$\frac{x_1+x_2+\cdots+x_n}n\ge(x_1x_2\cdots x_n)^{\frac1n},$$ with equality only when $x_1=x_2=\cdots=x_n.$

Proof. Let $g=(x_1x_2\cdots x_n)^{\frac1n}.$ According to the lemma we have $$\frac{x_1}g+\frac{x_2}g+\cdots+\frac{x_n}g\ge n,$$ i.e., $$\frac{x_1+x_2+\cdots+x_n}n\ge g=(x_1x_2\cdots x_n)^{\frac1n}.$$ Equality holds only if $\frac{x_1}g=\frac{x_2}g=\dots=\frac{x_n}g=1,$ that is, $x_1=x_2=\cdots=x_n.$

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    $\begingroup$ I enjoyed this proof a lot. There were so many nice elements. I like how the product is always positive, and how nicely the a_1a_2 terms cancel each other out when you add the inequalities, and how elegantly this lemma is applied to prove the problem. Did you come up with this on your own? $\endgroup$ – user230452 Feb 21 '16 at 7:10
  • $\begingroup$ How did you get step (2)? Shouldn't it be $a_1+a_2+a_3+\cdots$? $\endgroup$ – ghosts_in_the_code Oct 9 '16 at 3:52
  • $\begingroup$ Oh ok, u've considered $a_1a_2$ as a single number, not as two distinct ones. Got it. $\endgroup$ – ghosts_in_the_code Oct 9 '16 at 11:33
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Bernoulli's Inequality says that for $u\ge-1$ and $0\le r\le1$, $$ (1+u)^r\le1+ru\tag{1} $$ Setting $u=\frac xy-1$ in $(1)$ says that for $x,y\gt0$, $$ \left(\frac xy\right)^r\le(1-r)+r\frac xy\tag{2} $$ If we multiply $(2)$ by $y$, we get $$ x^ry^{1-r}\le rx+(1-r)y\tag{3} $$ Now $(3)$ can be used inductively to get $$ x_1^{r_1}x_2^{r_2}x_3^{r_3}\dots x_n^{r_n}\le r_1x_1+r_2x_2+r_3x_3+\dots+r_nx_n\tag{4} $$ where $r_1,r_2,r_3,\dots,r_n\ge0$ and $r_1+r_2+r_3+\dots+r_n=1$.

Inductive step:

Suppose that $(4)$ holds, then we can use $(3)$ to get $$ \begin{align} &\left(x_1^{r_1}x_2^{r_2}x_3^{r_3}\dots x_n^{r_n}\right)^{1-r_{n+1}}x_{n+1}^{r_{n+1}}\\ &\le(1-r_{n+1})\left(r_1x_1+r_2x_2+r_3x_3+\dots+r_nx_n\right)+r_{n+1}x_{n+1}\tag{5} \end{align} $$ where $(1-r_{n+1})(r_1+r_2+r_3+\dots+r_n)+r_{n+1}=1$

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There is a more direct induction proof.

We can assume that $0<a_1\leq \ldots \leq a_n$. Let set $A=\frac{a_1+ \ldots +a_n}{n}$. First notice that $(a_n-A)(A-a_1)\geq 0$, it can be rewritten as $\frac{a_1a_n}{A}\leq a_1+a_2-A $. Now we assume that the property hold for $n-1$. We can apply it with ${a_2,\ldots , a_{n-1},a_1+a_n-A}$. It gives : $$\left(\frac{\prod_{k=1}^na_k}{A}\right)^{1/(n-1)}\leq \left[(\prod _{k=2}^{n-1}a_k)(a_1+a_n-A)\right]^{1/(n-1)}\leq \frac{a_1+\ldots +a_n-A}{n-1}= A.$$

So : $$\prod_{k=1}^na_k\leq (A^{1+1/(n-1)})^{n-1}=A^n.$$

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This is a proof using Buffalo way (see this link for what buffalo way is http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=522084), I couldn't find it anywhere, I jush heard it's possible, so I tried it, hopefully there are no mistakes, but even if there are I think they will be easy to correct, the main idea should still be right. We'll proceed by strong induction, base case holds trivially from inequality $(x_1 - x_2)^2 \geq 0$, assume that AM-GM holds for all $k$ such that $1 < k < n$. Then we wish to prove that $$x_1^n + x_2^n + \cdots + x_n^n - nx_1x_2\cdots x_n \geq 0$$ This inequality is clearly symmetric, hence we may WLOG suppose $x_1 = \min\{x_1,x_2,\cdots,x_n\} = x$, therefore there exist $y_1,y_2,\cdots,y_{n - 1} \geq 0$ such that $x_i = x + y_{i - 1}$ for all $i \in \{1,2,\cdots,n\}$. Therefore the inequality can be rewritten as $$x^n + (x + y_1)^n + (x + y_2)^n + \cdots + (x + y_{n - 1})^n - nx(x + y_1)(x + y_2) \cdots (x + y_{n - 1}) \geq 0$$ This is a polynomial in $x$ and expanding this polynomial we get that the coefficient of $x^{n - k}$, where $1 < k < n$, is $$\binom{n}{k}p_k(y_1,y_2,\cdots,y_{n - 1}) - ne_k(y_1,y_2,\cdots,y_{n - 1})$$ where $p_k$ is $k$-th power sum and $e_k$ is $k$-th elementary symmetric polynomial. So it suffices to prove that $$\binom{n}{k}p_k(y_1,y_2,\cdots,y_{n - 1}) - ne_k(y_1,y_2,\cdots,y_{n - 1}) \geq 0$$ But by inductive hypothesis we get $$\frac{n}{k} \cdot \left(y^k_{i_1} + y^k_{i_2} + \cdots + y^k_{i_k}\right) - ny_{i_1}y_{i_2} \cdots y_{i_k} \geq 0,$$ where $i_1,i_2,\cdots,i_k \in \{1,2,3,\cdots,n - 1\}$ are pairwise distinct. Doing this for all the possible combinations of indices $i_1,i_2, \cdots, i_k$ we in fact get stronger inequality $$\frac{n}{k} \cdot \binom{n - 2}{k - 1} p_k(y_1,y_2,\cdots,y_{n - 1}) - ne_k(y_1,y_2,\cdots,y_{n - 1}) \geq 0.$$ Now also clearly coefficients of $x^n$ and $x^{n - 1}$ are $0$ and coefficient of $x^0$ is just $p_n(y_1,y_2,\cdots,y_{n - 1})$. Hence all the coefficients of our polynomial are non-negative, therefore the polynomial is non-negative, thus the inductive step is proved and the whole proof is finished.

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Draw graph of $y= \ln x$.

Choose $n$ points on it.A polygon of $n$ sides is thus formed whose coordinates of whose centroid is $$G \left (\frac{(x_1+x_2+....x_n)}n, \frac{(y_1 + .....y_n)}n\right ) = \left (\frac{(x_1+x_2+....x_n)}n, \frac{(\ln x_1 + \ln x_2 + .... + \ln x_n)}n\right )$$ Form centroid we draw a perpendicular to $x$ axis and extend it to meet the graph of $y= \ln x$ at $C$ and $y$ axis at $D$.

Now coordinates of point $C$ are $\left (\frac{(x_1+x_2+....x_n)}n, \ln \left (\frac{x_1+x_2+....x_n}n\right )\right )$. Now $$CD= \ln {\left (\frac{x_1+x_2+....x_n}n\right )}$$ ($ y$ coordinate of $C$). $$GD = \frac{\ln x_1 + \ln x_2 + .... + \ln x_n}n = \frac{\ln (x_1\cdot x_2\cdot ....x_n)} n$$ ($y$ coordinate of $G$). Now $CD> GD$ (as centroid of a convex polygon is always inside the polygon) $$\therefore \ln {\left (\frac{x_1+x_2+....x_n}n\right )} > \frac{\ln (x_1\cdot x_2\cdot ....x_n)} n = \ln \left ((x_1 \cdot x_2 \cdot ... x_n)^{\frac 1n} \right )$$ Now base and the number are positive. So we take antilog of both sides and get$$\frac{x_1+x2+....x_n}n > (x_1\cdot x_2\cdot ....x_n)^{1/n}$$ Now, it is also possible that all the selected $n$ points are only one point. Hence then the centroid is the point itself. Hence $\frac{x_1+x_2+....x_n}n = (x_1\cdot x_2\cdot ....x_n)^{1/n}$. Thereby proving that $\frac{x_1+x_2+....x_n}n \geq (x_1\cdot x_2\cdot ....x_n)^{1/n}$, where equality holds when$ x_1=x_2=...= x_n$.

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  • $\begingroup$ Can you please fix the formatting. $\endgroup$ – user99914 Jun 7 '15 at 3:33
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This is not an original or different proof but since any question, asking for a proof for this inequality, gets closed as a duplicate of this question ... I feel this answer should be here too.


First, let me prove the result that if there are two numbers of the same sum, the numbers which are closer together have a greater product.

$$ t + u = x + y $$

Let $t,u$ be further apart from each other than $x,y$

But, $$ t - u \gt x - y $$

Consider the following identity, writing the product of two numbers as a function of it's sum and difference. Since we are subtracting a positive quantity, and the sum is the same, the greater product is of the numbers with the lesser difference.

$$4ab = (a+b)^{2} - (a-b)^{2}$$

From this, we clearly see that $$xy \gt tu$$

Now, let's consider a set of numbers

$$a, b, c {\dots} n$$

Let us choose its arithmetic mean, $M$ given by

$$M = \frac{a+b+c+{\dots}n}{n}$$

The product of this series is

$$P = a.b.c.{\dots}n$$

Now, we make another series of numbers $$a_{1}, b_{1}, c_{1} {\dots} n_{1}$$

We choose this series in such a way that all elements are equal but we make one change. We choose one number less than $M$, (say $a_{1}$) and write $a_{1} = M$.

However, we'd like to preserve the sum and the $AM$ of this series so we'd choose a number greater than $M$, say $b_{1}$ and reduce its value so that $a_{1} + b_{1} = a + b$. Since, $a_{1}$ has been increased, $b_{1}$ must be decreased. All the other values remain as their older counterparts.

$c_{1} = c, d_{1} = d, {\dots} n_{1} = n$

Now we observe,

$P_{1} = a_{1}.b_{1}.c_{1}{\dots}n_{1}$ $=Mb_{1}.c{\dots}n$

Since the product of the terms $c_{1}d_{1}{\dots}n_{1} = cd{\dots}n$, and $Mb_{1} \gt ab$,

$$\implies P_{1} \gt P$$

Now, we construct another series of terms, $a_{2}, b_{2},c_{2}{\dots}n_{2}$ in the same way where all the terms are equal to their counterparts of the previous series. We write, $ b_{2} = M$, and increase the value of a number less than $M$, (say $c_{2}$) so that the sum $b_{2}+c_{2} = b_{1} + c_{1}$ is preserved, but $b_{2}c_{2} \gt b_{1}c_{1}$, since $b_{2},c_{2}$ are closer together.

Now, we observe the product of this series.

$$P_{2}= a_{2}b_{2}c_{2}{\dots}n_{2}$$ $$\implies P_{2} = M^{2}c_{2}{\dots}n_{2}$$

So, by the similar logic, $$P_{2} \gt P_{1}$$

We go on constructing many series making a new element equal to M each time and increasing the product we get,

$$P_{n} \gt {\dots} \gt P_{2} \gt P_{1} \gt P$$ $$\implies P_{n} = M^{n} \gt P$$ $$\implies ( {\frac{a+b+c+\dots+n}{n}})^{n} \gt {a.b.c.{\dots}n}$$

But, both the $L.H.S.$ and the $R.H.S$ are positive quantities.

$$\implies \frac{a+b+c\dots+n}{n} \gt {(a.b.c.{\dots}n)}^{\frac{1}{n}}$$

And, Voila !

$$\implies A.M \gt G.M.$$

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Let $S(n)$ denote the statement $$ S(n):\; \frac{x_1+x_2+\cdots+x_n}{n}\geq\sqrt[n]{x_1x_2\ldots x_n},\quad n\in\mathbb{N}. $$ Base step ($n=1$): The statement $S(1)$ says that $\frac{x_1}{1}\geq\sqrt[1]{x_1}$, which is true because $x_1 = x_1$.

Base step ($n=2$): The statement $S(2)$ says that $$ \frac{x_1+x_2}{2}\geq\sqrt{x_1x_2},\tag{1} $$ which is true because $$ a\leq x \leq b \longleftrightarrow a+b\geq x+\frac{ab}{x}, \qquad 0<a\leq b,\; x>0 $$

Inductive step ($S(k)\to S(k+1)$): Fix some $k\geq 1$, where $k\in\mathbb{N}$. Assume that $$ S(k):\; \frac{x_1+x_2+\cdots+x_k}{k} \geq \sqrt[k]{x_1x_2\ldots x_k} $$ holds. To be proved is that $$ \frac{x_1+x_2+\cdots+x_{k+1}}{k+1}\geq\sqrt[k+1]{x_1x_2\ldots x_{k+1}} $$ follows. If $x_1 = x_2 = \cdots = x_{k+1}$, then the proof is done. If not, let $x_1x_2\ldots x_{k+1} = \rho^{k+1}$. Without loss of generality, assume that $x_1\leq x_i$ and $x_i \leq x_2$ for all $i$; that is, assume that $x_1 < \rho < x_2$. Beginning with the left side of $S(k+1)$ [excluding the $k+1$ divisor], \begin{align} x_1+x_2+\cdots+x_{k+1} &> \rho+\frac{x_1x_2}{\rho}+x_3+\cdots+x_k+x_{k+1}\tag{by $(1)$}\\ &\geq \rho+k\cdot\left(\sqrt[k]{\frac{x_1x_2}{\rho}x_3\ldots x_{k+1}}\right)\tag{by $S(k)$}\\ &= (k+1)\rho, \end{align} one arrives at the right side of $S(k+1)$ [with a $k+1$ multiple], thereby showing that $S(k+1)$ is also true, completing the inductive step. Thus, by mathematical induction, $S(n)$ is true for all $n\geq 1$, where $n\in\mathbb{N}$.

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Just consider a probabilistic approach of the proof using the convexity of the function $-\log x$ and using Jensen's inequality.

Let $\{a_1,a_2,a_3,\ldots,a_n\}$ be a set of positive numbers. Consider a distribution for a random variable $X$ by placing weight $1/n$ on each of these numbers.Then the mean of the random variable $X$ is the arithmetic mean (AM), $$E(X)=n^{-1}\sum_{i=1}^n a_i$$Then since $-\log x$ is a convex function,we have by Jensen's inequality that

$$-\log\left(\frac {\sum_{i=1}^n a_i}{n}\right)\le E(-\log X)=-\frac{1}{n}\left(\sum_{i=1}^n \log a_i\right)=-log(a_1a_2a_3....a_n)^{1/n}$$ or equivalently $$\log\left(\frac {\sum_{i=1}^n a_i}{n}\right)\ge \log(a_1a_2a_3....a_n)^{1/n}$$ and hence we can see that $$(a_1a_2a_3....a_n)^{1/n}\le \frac{\sum_{i=1}^n a_i}{n}$$ As we can see the inequality on the left hand side of this inequality is the geometric mean and the right hand side is the arithmetic mean for any set of positive numbers. Note: If you replace $a_i$ by $1/a_i$ we get the relationship between the harmonic mean ,the arithmetic mean and the geometric mean

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  • $\begingroup$ my favorite proof. utilizing the fact that this is a corollary of the behavior of convex functions and tying it together is much better IMO than ad hoc algebraic manipulations $\endgroup$ – Charlie Tian Mar 9 '18 at 7:49
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Diagram

In the diagrams, the circle is tangential to the horizontal line at $C$ and intersects the vertical line at $A$ and $B$, while the lines themselves intersect at $P$. In the figure on the left, let $a = PA$, $b = PB$, and $c = PC$. By the power of the point $P$, we know $c^2 = ab$ or $c = \sqrt{ab}$.

Then, we shift the circle to the right so that it is tangential to both lines. The circle being a symmetric figure, we know the new $PA$ is $\frac{a + b}{2}$ and therefore $PC$ is also $\frac{a + b}{2}$. But as we moved the circle to the right, the new $PC \geq c$ and hence $$\frac{a + b}{2} \geq \sqrt{ab}$$ (Note that equality occurs when $A$ and $B$ are the same in the figure on the left.)

I am not sure about this, but I think the proof can be extended to the AM-GM inequality with 3 variables if we consider a sphere with one tangential plane and one intersecting plane, and we apply the above proof to each 'slice'... by extension, one could consider the $n$-dimensional sphere intersecting with $(n-1)$-dimensional planes, but as I said, I am very unsure about this.

(The credit for this proof goes to Sujay Kazi, a friend at a math club I attend. He recently presented this proof and I cannot recall seeing it in the usual lists of AM-GM proofs, so I am including it here.)

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  • $\begingroup$ Very nice proof! :) $\endgroup$ – Sandeep Silwal 2 days ago
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Applying the Log sum inequality (direct consequence of Jensen inequality)

$$ \sum a_i \log \frac{ b_i}{a_i} \le a \log \frac{b}{a}$$ where $a=\sum a_i$ and $b=\sum b_i$, $a_i\ge 0$, $b_i\ge 0$; setting $b_i=x_i/n$ $a_i=1/n$ we get

$$ \frac{1}{n}\sum \log x_i \le \log \left( \frac{\sum x_i}{n}\right) $$ or

$$\log\left (\frac{x_1+ \ldots + x_n}{n} \right) \geq \frac{ \log x_1 +\log x_2 +\cdots \log x_n}{n}$$

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This is a plausibility argument rather than a complete proof. Without loss of generality we may assume that $x_1,x_2,\dots,x_n\in[0,1].$ Let $p_i=x_i^{1/n}\in[0,1].$

Consider the following variant of Russian Roulette. You are faced with a panel of $n$ buttons. If you press the $i^{\text{th}}$ button, then with probability $p_i$ you will get a fatal electric shock. You are required to press a button $n$ times. The probabilities $p_1,\dots,p_n$ are known to you, but the buttons are not labeled, you don't know which is which. You can choose between two strategies.

I. Press each button once.

II. Choose a button at random and press it $n$ times.

It is INTUITIVELY OBVIOUS that strategy II is better: you have a better chance of survival pressing the same button that you already pressed and lived, than pressing an untried button. (Prove this rigorously and you have a proof of the AM-GM inequality.)

Now, the probability of survival with strategy I is $$p_1p_2\cdots p_n=(x_1x_2\cdots x_n)^{1/n}$$ and with strategy II it's $$\frac{p_1^n+p_2^n+\cdots+p_n^n}n=\frac{x_1+x_2+\cdots+x_n}n,$$ so the fact that strategy II dominates strategy I is expressed by the AM-GM inequality $$(x_1x_2\cdots x_n)^{1/n}\le\frac{x_1+x_2+\cdots+x_n}n.$$

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$$ \frac{x_1+x_2+....+x_n}{n}\geq\sqrt[n]{x_1x_2....x_n} $$

Using $\color{blue}{\text{Forward-Backward Induction/Cauchy Induction}}$,

Our base case is $n=2$ as $n=1$ is trivial.

Base Case: $$ P(2):(\sqrt{x_1}-\sqrt{x_2})^2\geq0\implies x_1+x_2-2\sqrt{x_1x_2}\geq0\\ \implies x_1+x_2\geq2\sqrt{x_1x_2}\implies \frac{x_1+x_2}{2}\geq\sqrt{x_1x_2} $$ $P(2)$ is true.

Inductive Step:

1.) Forward Part:

Let the inequality holds for $n=k$, ie. $P(k)$ is true. $$ P(k):\frac{x_1+x_2+...+x_k}{k}\geq\sqrt[k]{x_1.x_2....x_k} $$ $$ P(2k):\frac{x_1+x_2+...+x_{2k}}{2k}=\frac{1}{k}\Big(\frac{x_1+x_2}{2}+\frac{x_3+x_4}{2}+...+\frac{x_{2k-1+x_{2k}}}{2}\Big)\geq\frac{\sqrt{x_1x_2}+\sqrt{x_3x_4}+...+\sqrt{x_{2k-1}x_{2k}}}{k}\\\geq\sqrt[k]{\sqrt{x_1x_2x_3....x_{2k-1}x_{2k}}}=\sqrt[2k]{x_1x_2...x_{2k}} $$ $P(2k)$ is true whenever $P(k)$ is true.

2.) Backward Part:

Substitute $x_k=\frac{x_1+x_2+...+x_{k-1}}{k-1}$ in $P(k)$, $$ P(k-1):\frac{x_1+x_2+...+x_k}{k}\geq\sqrt[k]{x_1x_2...x_k}\\ \frac{x_1+x_2+...+x_{k-1}+\frac{x_1+x_2+...+x_{k-1}}{k-1}}{k}\geq\sqrt[k]{x_1x_2...x_{k-1}.\frac{x_1x_2...x_{k-1}}{k-1}}\\ \bigg(\frac{x_1+x_2+...+x_{k-1}}{k-1}\bigg)^k\geq x_1x_2...x_{k-1}.\frac{x_1x_2...x_{k-1}}{k-1}\\ \bigg(\frac{x_1+x_2+...+x_{k-1}}{k-1}\bigg)^{k-1}\geq\frac{x_1x_2...x_{k-1}}{k-1}\\ \frac{x_1+x_2+...+x_{k-1}}{k-1}\geq \sqrt[k-1]{\frac{x_1x_2...x_{k-1}}{k-1}} $$ $P(k-1)$ is true whenever $P(k)$ is true.

$\implies$ By the forward-backward induction AM-GM inequality is true for all $n$.

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If $a_1,...,a_n>0$, and $0<p\le q$, by Hölder inequality: $$\left(\frac{1}{n}\sum_{k=1}^na_k^p\right)^{\frac{1}{p}}\le\left(\frac{1}{n}\sum_{k=1}^na_k^q\right)^{\frac{1}{q}}.$$ Also, using De l'Hopital we get: $$\lim_{p\rightarrow 0^+}\left(\frac{1}{n}\sum_{k=1}^na_k^p\right)^{\frac{1}{p}}=\exp\left(\frac{1}{n}\sum_{k=1}^n \log(a_k)\right)=\left(\prod_{k=1}^na_k\right)^{\frac{1}{n}}$$ So: $$\forall q>0, \left(\prod_{k=1}^na_k\right)^{\frac{1}{n}}\le\left(\frac{1}{n}\sum_{k=1}^na_k^q\right)^{\frac{1}{q}}.$$ In particular for $q=1$ we get AM-GM.

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  • $\begingroup$ I think Hölder's inequality is way more general than AM-GM and proving the latter through the former is something artificial. $\endgroup$ – pointguard0 Aug 4 '18 at 0:56
  • $\begingroup$ @pointguard0 If that is your problem, you can always use Cauchy Schwarz to get the result for $p=1/2,q=1$, then $p=1/4, q=1/2$, and so on... $\endgroup$ – Bob Aug 4 '18 at 4:43
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Just came up with this proof that uses only the Cauchy-Schwarz inequality. Not sure if it's already known.

We first show the following result for polynomials.

Let $p$ be a polynomial with non-negative coefficients. Then $\sqrt{p(a) \, p(b)} \geq p(\sqrt{ab})$ for all $a, b \geq 0$.

Proof. Let $p(x) = c_n x^n + c_{n-1} x^{n-1} + \cdots + c_0$. Then let $\vec{v} := (\sqrt{c_n a^n}, \sqrt{c_{n-1} a^{n-1}}, \dots, \sqrt{c_0 a^0})$, and similarly define $\vec{w} = (\sqrt{c_n b^n}, \sqrt{c_{n-1} b^{n-1}}, \dots, \sqrt{c_0 b^0})$. The result is immediate from Cauchy-Schwarz. $\square$

We use this result to show AM-GM for two variables.

Let $a, b \geq 0$. Then $\frac{a+b}{2} \geq \sqrt{ab}$.

Proof. Let $P_n$ be the $n^\text{th}$-order Taylor polynomial of $e^x$. By the previous result, we have $$\sqrt{P_n(a) \, P_n(b)} \geq P_n(\sqrt{ab})$$ for all $n \geq 0$. The desired result follows by taking $n \to \infty$ and noting that $e^x$ is increasing. $\square$

I am not sure whether this result can be extended to multiple variables.

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My idea is to use only the Cauchy Schwartz inequality by induction.

Applying the Cauchy_Schwartz inequality on the vectors $(\sqrt{a},\sqrt{a})^T$ and $(\sqrt{b},\sqrt{b})^T$ we get: $$\sqrt{ab} +\sqrt{ab} \le (a+b)^{1/2}(a+b)^{1/2} $$ that is $$\sqrt{ab} \le \frac{a+b}{2}~~~~~~\color{red}{\text{AM-GM for $n =2$}}$$

Hypothesis of induction: Assume that for all $p\in\{1,2,\cdots, n-1\}$ we have, \begin{align*} \color{blue}{(a_1a_2\cdots a_p)^{1/p}\leq \frac{a_1+a_2+\cdots+a_p}{p}} \end{align*} We want to prove that it is true for $n$. we will discuss two case: $n =2p$ and $n=2p-1$

If $n =2p$ Then we proceed as follows: Using that case: $n=2$ and the case $n=p$, we have,

$$ \begin{align}\color{blue}{\frac{a_1+a_2+\cdots+a_n}{n} }&= \frac{1}{2}\left(\overbrace{\frac{a_1+a_2+\cdots+a_p}{p}}^a+\overbrace{\frac{a_{p+1}+a_{p+2}+\cdots+a_n}{p}}^b \right) \\ &\ge\sqrt{\left(\frac{a_1+a_2+\cdots+a_p}{p}\right) \left(\frac{a_{p+1}+a_{p+2}+\cdots+a_n}{p}\right)}\\ &\ge\sqrt{(a_1a_2\cdots a_p)^{1/p}(a_{p+1}a_{p+2}\cdots a_n)^{1/p}}\\ &=\sqrt{(a_1a_2\cdots a_n)^{1/p}} = \color{blue}{(a_{p+1}a_{p+2}\cdots a_n)^{1/n} } ~~~~~~\color{red}{\text{AM-GM for $n =2p$}} \end{align}$$

If $n =2p-1$ (we use the case $n=2p$, as follows.we have, $n+1 =2p $ and $p<n$ : Then taking $$ a_{n+1} = \frac{a_1+a_2+\cdots+a_n}{n}$$ in the previous case we obtain,

\begin{align}&\frac{a_1+a_2+\cdots+a_{n+1}}{n+1} \ge (a_1a_2\cdots a_{n+1})^{\frac{1}{n+1}}\\ &\Longleftrightarrow\frac{1}{n+1}\left(a_1+a_2+\cdots+a_n+\overbrace{\frac{a_1+a_2+\cdots+a_n}{n}}^{a_{n+1}} \right) \ge \left(a_1a_2\cdots a_{n}\left(\overbrace{\frac{a_1+a_2+\cdots+a_n}{n}}^{a_{n+1}}\right) \right)^{\frac{1}{n+1}}\\ \\&\Longleftrightarrow\frac{a_1+a_2+\cdots+a_n}{n} \ge \left(a_1a_2\cdots a_{n}\left(\frac{a_1+a_2+\cdots+a_n}{n}\right) \right)^{\frac{1}{n+1}}\\ &\Longleftrightarrow\left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^{\frac{n}{n+1}} \ge \left(a_1a_2\cdots a_{n}\right)^{\frac{1}{n+1}} \\&\Longleftrightarrow \color{blue}{\frac{a_1+a_2+\cdots+a_n}{n} \ge \left(a_1a_2\cdots a_{n}\right)^{\frac{1}{n}}}~~~~\color{red}{\text{AM-GM for $n =2p-1$}}\end{align}

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