Is there a proof of the AM-GM Inequality via Cauchy's Inequality? Or vice-versa?

The title says it all. There is already a list exhausting proofs of the AM-GM inequality here; however, none via Cauchy's Inequality.

Why do I care? Recreation. But more than that, proving inequalities via each other is one of the predominant past-times in the theory of inequalities. I'm slightly surprised this isn't a common proof out there. And a proof via this route sounds like it would demonstrate either extreme algebraic subtlety and cleverness or give rise to a new technique in proving inequalities. Venerable ends, each of them.

Undoubtedly, Cauchy's Inequality and the AM-GM Inequality are connected in some fashion. Jensen's Inequality can prove both of them. However, it is interesting to note that difference in the proofs via Jensen's. The proof of Cauchy via Jensen relies on the boring observation that $$x^2$$ (an algebraic function) is strictly convex followed by algebraic hyjinx and cleverness in choosing weights and evaluation points. On the other hand, the proof of the AM-GM Inequality via Jensen relies more on the foresight or prescience of using the exponential function (a transcendental function) with no funny business involving the weights or evaluation points.

Both can be proved by Cauchy Induction, although the base cases differ in that they rely on $$\sqrt{xy}\leq \frac{x+y}{2}\quad\text{and}\quad(x_1y_1+x_2y_2)^2\leq (x_1^2+x_2^2)(y_1^2+y_2^2)$$ respectively. Both can be proved by normalization, although the exact reductions they are proved from differ. Cauchy relies on $$\sum_{k=1}^n |x_ky_k|\leq 1\quad\text{whenever}\quad\sum_{k=1}^nx_k^2=1=\sum_{k=1}^ny_k^2$$ while the AM-GM relies on $$\alpha_1+\alpha_2+\cdots+\alpha_n\geq n\quad\text{whenever}\quad \alpha_1\alpha_2\cdots\alpha_n=1\,.$$

Frequently, one can discover that one inequality can be proved from another if they have similar sets where equality can hold. For example, Minkowski's inequality is frequently proven with Holder's Inequality and the triangle inequality (an inequality which is typically too coarse for most things). However, the triangle inequality and Minkowski's Inequality share very similar sets where equality can happen. This makes me focus on the points where equality can hold in either inequality.

Let $$E=(0,\infty)$$. Then the set where equality can hold in Cauchy's inequality is $$\{(x;y)\in E^{2n}:\text{ there is a }\lambda>0\text{ such that }x=\lambda y\}$$ while the set where equality can hold in the AM-GM inequality is $$\{x\in E^n: \text{there is a }\lambda>0 \text{ such that } x=\lambda(1, 1, \ldots, 1)\}\,.$$ These are extremely different sets and hint that a proof between the two might be difficult to find. However, I doubt it's impossible. There will undoubtedly be trickery involved in passing between these different dimensions.

To lend some credence that it could be possible, Cauchy's Inequality is strong enough to prove the AM-HM Inequality for it is clear that $$n^2\leq \left(\sum x_k\right)\left(\sum 1/x_k\right)\,.$$ This proof requires a clever choice of '$$y$$' in terms of '$$x$$' to pass to the lower dimension. Thus, it might be possible to find a function $$f$$ such that if $$x=(x_1, x_2, \ldots, x_n)$$ and $$y=(f(x_1), f(x_2), \ldots, f(x_n))$$ then Cauchy's Inequality could result in the AM-GM inequality.

• You can, by proving AMGM for 2 variables and then extending it to 3,4,... See here: math.stackexchange.com/questions/678174/… Commented Nov 15, 2017 at 5:03
• Cauchy's inequality and the AMGM inequality agree for two arguments (with a square root difference). To use a simple case of the Cauchy inequality to use in Cauchy induction and say the AMGM was proved by the former, although technically correct, seems cheap and uninteresting. The proof by factorization in your link seems interesting. But this begs the question of what you would multiply by in order to recover the AMGM inequality.
– user123641
Commented Nov 15, 2017 at 5:11
• Let $H(t)=((x_1^t+\cdots +x_n^t)/n)^{1/t}$ be the $t$th power mean. Cauchy's Inequality can readily show that $H(1)\geq H(1/2)\geq H(1/4)\geq H(1/8)\geq\cdots$. And it is provable that $\lim_{t\rightarrow 0} H(t)$ is the geometric mean of $x_1, \ldots, x_n$. This, in a sense, passes the buck however. This limit is typically proven with the exponential or logarithmic function. And proving that any one of the $H(1/2^m)$ is greater than the geometric mean is no easier than directly proving the AM-GM from Cauchy.
– user123641
Commented Nov 16, 2017 at 1:00
• @Robert It seems the answers below does not satisfy your quires . if not why did you chose to waste your bounty? Commented Nov 27, 2017 at 13:04

A nice proof can be found in chapter 2 of The Cauchy-Schwarz Master Class by J.M. Steele. It is stated in form of some problems starting with:

Problem 2.1: Show that for every sequence of nonnegative real numbers $a_1,a_2,\ldots,a_n$ one has \begin{align*} (a_1a_2\cdots a_n)^{1/n}\leq \frac{a_1+a_2+\cdots a_n}{n}\tag{1} \end{align*}

The proof of this and the following problems is done in some steps.

Step 1: $n=2^k, k\geq 1$

From the Cauchy-Schwarz inequality \begin{align*} \sqrt{a_1a_2}\leq \frac{a_1+a_2}{2}\tag{2} \end{align*}

which is (1) with $n=2$ we can go on one step to $n=4$ by applying (2) twice \begin{align*} (a_1a_2a_3a_4)^{\frac{1}{4}}\leq \frac{(a_1a_2)^{\frac{1}{2}}+(a_3a_4)^{\frac{1}{2}}}{2} \leq \frac{a_1+a_2+a_3+a_4}{4} \end{align*}

This new bound can be used to prove (1) for $n=8$ \begin{align*} (a_1a_2\cdots a_8)^{\frac{1}{8}}\leq \frac{(a_1a_2a_3a_4)^{\frac{1}{4}}+(a_5a_6a_7a_8)^{\frac{1}{4}}}{2} \leq \frac{a_1+a_2+\cdots+a_8}{8} \end{align*}

Iteratively going on this way we can show AM-GM for $n=2^k,k\geq 1$.

\begin{align*} \color{blue}{(a_1a_2\cdots a_{2^k})^{\frac{1}{2^k}}\leq\frac{a_1+a_2+\cdots+a_{2^k}}{2^k}} \end{align*}

Step 2: $2^{k-1}<n<2^k$

The next step is to fill the gaps between $2^{k-1}$ and $2^k$. This is done by taking a value $n<2^k$, considering $\alpha_i=a_i$ for $1\leq i\leq n$ and setting \begin{align*} \alpha_i=\frac{a_1+a_2+\cdots+a_n}{n}\equiv A\qquad\qquad n<i\leq 2^k \end{align*}

The sequence $\{a_i:1\leq i\leq n\}$ is transformed to a sequence $\{\alpha_i:1\leq i\leq 2^k\}$ padded with $2^k-n$ copies of the average $A$ and it can be shown \begin{align*} \color{blue}{(a_1a_2\cdots a_n)^{1/n}\leq \frac{a_1+a_2+\cdots+a_n}{n}} \end{align*}

Step 3: rational numbers

Based upon step (2) the AM-GM is shown for non-negative rational numbers $p_1,p_2,\ldots,p_n$ that sum to one. \begin{align*} \color{blue}{a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}\leq p_1a_1+p_2a_2+\cdots +p_na_n} \end{align*}

Step 4: real numbers

Based upon step (3) the AM-GM is shown for non-negative real numbers $p_1,p_2,\ldots,p_n$ that sum to one by taking limits.

\begin{align*} \color{blue}{a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}\leq p_1a_1+p_2a_2+\cdots +p_na_n} \end{align*}

Note: All these steps are some kind of bootstrapping based upon the step before and starting with the Cauchy-Schwarz inequality.

• Cauchy's inequality and the AMGM inequality agree for two arguments (with a square root difference). To use a simple case of the Cauchy inequality to use in Cauchy induction and say the AMGM was proved by the former, although technically correct, seems cheap and uninteresting.
– user123641
Commented Nov 19, 2017 at 5:26
• I own the book. And this was one of the proofs that I briefly mentioned in the question.
– user123641
Commented Nov 19, 2017 at 5:29
• @Robert: Sorry, that the answer does not please you. To me it seems not uninteresting to me it seems standard. Commented Nov 19, 2017 at 7:12

Here We use only Cauchy Schwartz inequality by induction.

Since $$a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b})^2\geq 0$$ we get $$\sqrt{ab} \le \frac{a+b}{2}~~~~~~\color{red}{\text{AM-GM for n =2}}$$ Now for a better understanding of what is going to happen in the sequel, let us prove for $$n= 4$$ and then for $$n= 3$$

Using the case: $$n=2$$ twice, we have, $$\color{blue}{\frac{x+y+z+t}{4} =} \frac{1}{2}\left(\overbrace{\frac{x+y}{2}}^a+\overbrace{\frac{z+t}{2}}^b \right) \ge\sqrt{\left(\frac{x+y}{2}\right) \left(\frac{z+t}{2}\right)}\\\ge\sqrt{\sqrt{xy} \sqrt{zt}} = \color{blue}{\sqrt[4]{xyzt} } ~~~~~~\color{red}{\text{AM-GM for n =4}}$$ this proves the result for $$n=4$$

Now taking $$t=\frac{x+y+z}{3}$$ in the above inequality, we get,

\begin{align}\frac{x+y+z+t}{4} \ge \sqrt[4]{xyzt}&\Longleftrightarrow\frac{1}{4}\left(x+y+z+\overbrace{\frac{x+y+z}{3}}^t \right) \ge \left(xyz\left(\overbrace{\frac{x+y+z}{3}}^t\right) \right)^{1/4}\\ \\&\Longleftrightarrow\frac{x+y+z}{3} \ge \left(xyz\left(\frac{x+y+z}{3}\right) \right)^{1/4}\\ &\Longleftrightarrow\left(\frac{x+y+z}{3}\right)^{3/4} \ge \left(xyz\right)^{1/4} \\&\Longleftrightarrow \color{blue}{\frac{x+y+z}{3} \ge \sqrt[3]{xyz}}~~~~\color{red}{\text{AM-GM for n =4}}\end{align}

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)}\\ &\geq\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_{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 $$p : Then taking $$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}

Just came up with this proof. 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. $$\square$$

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.

• Deserves an upvote just for the novelty. But your argument does readily extend for arguments of powers of 2, i.e., $$\sqrt[2^n]{p(x_1)\cdots p(x_{2^n})}\leq p(\sqrt[2^n]{x_1\cdots x_{2^n}}).$$ This gives the AMGM for powers of 2 by using the same polynomials and taking limits. The other cases can be recovered by Cauchy's classic argument. Closest answer so far.
– user123641
Commented Oct 6, 2019 at 16:18