# Geometric mean never exceeds arithmetic mean

This was a mathematical induction question proposed in a textbook, and I've exhausted multiple approaches (proving RHS - LHS > 0, splitting the fraction, fractional exponents, etc.)

The geometric mean of $n$ positive numbers $a_1, a_2,\ldots,a_n$ is $\sqrt[n]{a_1a_2 \ldots a_n}$ and their arithmetic mean is $\frac{a_1+a_2+\ldots+a_n}{n}$. If $a_1, a_2,\ldots,a_n$ are $n$ positive real numbers, prove by induction that their geometric mean is always smaller than or equal to their arithmetic mean, i.e. $\sqrt[n]{a_1a_2\ldots a_n} \leq \frac{a_1+a_2+\ldots+a_n}{n}$

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Not what you are looking for, but the Lagrange multipliers proof is pretty slick. Just take the geometric mean as the function to minimize and the constraint $\sum_{i=1}^n x_i=n$. – BBischof Mar 16 '11 at 5:14
For a big list, see Proofs of AM-GM inequality – barto Feb 20 '15 at 18:01

Hint:

Using induction:

1. Show the base case $n = 2$.
2. Show that if the statement is true for $2^n$, then it is true for $2^{n+1}$.
3. Show that if the statement is true for $n$, then it is true for $n - 1$.

(This method of induction is sometimes called Cauchy induction).

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Assuming true for $2^{k}$ and attempting to prove true for $2^{k+1}$, I end up with the inequality, $\frac{a_1 + a_2 + \ldots + a_k + a_{k+1}}{2^{k+1}} \geq \frac{1}{2}\Big((a_1a_2\ldots a_k)^{\frac{1}{2^k}}+\frac{a_{k+1}}{2^k}\Big)$ but, I have no idea how to proceed next, to reduce the RHS to $(a_1a_2\ldots a_{k+1})^{\frac{1}{2^{k+1}}}$? – amy Mar 16 '11 at 5:00
@Jaydon: Your indices seem to be wrong. You need to be finding the means of the terms $a_1$ through $a_{2^{k+1}}$, not $a_{k+1}$. In order to show the $2^{k+1}$ case, try taking the means of the means of $a_1,\ldots,a_{2^k}$ and $a_{2^k+1},\ldots,a_{2^{k+1}}$. – Alex Becker Mar 16 '11 at 5:17
@Alex: Thanks for the correction! – amy Mar 16 '11 at 5:20
@Alex: I don't really understand what you mean by your suggestion to take the "means of the means" and which terms to apply that to. Can you please elaborate? – amy Mar 16 '11 at 5:31
@Jaydon: take $mean(mean(a_0,\ldots,a_{2^k}),mean(a_{2^k+1},\ldots,a_{2^{k+1}}))$ using the arithmetic and geometric means and compare them. – Alex Becker Mar 16 '11 at 5:54

Hint: For induction, you need to prove the base case: $\sqrt{a_1+a_2} \le \frac{a_1+a_2}{2}$. If you square both sides... Then you need to prove that if it works for $n$ numbers, it works for $n+1$, so put the averages for $a_1 \ldots a_n$ in for $a_1$.

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An induction proof which I like, is to prove that

If $\displaystyle \prod_{j=1}^{n} a_j = 1$ then $\displaystyle \sum_{j=1}^{n} a_j \ge n$

Now for the induction step, if $\displaystyle \prod_{j=1}^{n+1} a_j = 1$, pick $\displaystyle a_r$ and $\displaystyle a_s$ such that $\displaystyle 0 \ge (1-a_r)(1-a_s)$ and...

For a different proof, one could claim that using the concavity of $\displaystyle \log x$ is an induction proof too (going from $\displaystyle t + (1-t) = 1$ to $\displaystyle t_1 + t_2 + \dots + t_n = 1$).

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Awwh, come on, folks. There are 12 (yes, 12) proofs of the GM-AM Inequality in the classic little book on inequalities by Beckenbach and Bellman.

Here is an excerpt from a reader’s review of the book at Amazon:

An Introduction to Inequalities is an unexpectedly delightful book. Relatively brief, only 129 pages, this publication of The Mathematical Association of America, requires no more than basic high school mathematics. Nonetheless, I am convinced that Edwin Beckenbach's and Richard Bellman's systematic study of inequalities would interest most students in an early calculus course.

Some classical inequalities were familiar, like the arithmetic mean - geometric mean inequality and the Cauchy inequality (two-dimensional version). But others like the n-dimensional version of the Cauchy inequality (along with the Cauchy-Lagrange identity), the Hölder inequality, and the Minkowski inequality were new to me. What I found most surprising was how these classical inequalities were so interrelated, and how some can be considered generalizations of others. Beckenbach and Bellman introduce clever substitutions to transform one inequality expression into another.

(credit: The reviewer quoted is Michael Wischmeyer, of Houston, Texas.) Here is the link.

There is an old joke that says that sometimes a couple of months spent in the laboratory can save a couple of hours spent in the library.

In other words, let’s refrain from re-inventing the wheel, except purely as an exercise.

Regards,

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Following JavaMan's hint, I came up with the following :

Let the statement$\:\:P(\eta)\equiv\large\frac{1}{\eta}\sum_\limits{j=1}^\eta\:{\normalsize a_j}\:\geq\:\large\prod_\limits{j=1}^\eta\sqrt[\large \eta] {\normalsize a_j}$ $\:\:\:\:\:\:\:\:\:\:\:\:\eta=2^j,\:\:j\in\mathbb{N}$

Now, $\:P(1)\equiv\large\frac{(a_1+a_2)}{2}$$\:\geq\:\sqrt {a_1a_2}\:\:\longleftrightarrow \:\:a_1-\small2$$\sqrt {a_1a_2}+a_2\:\geq\:0\:\:\longleftrightarrow \:\:\left(\sqrt {a_1}-\sqrt {a_2}\right)^2\:\geq\:0$
wich is certainly true.

Let suppose that$\:P(\eta)\:$holds$\:\forall\eta\:$and let's go and show that it also holds for $\:P(\eta+1)\:$

We have$\:\:2\eta=2^{j+1}\:\:\to \:\:\:\large\frac{1}{2\eta}\sum_\limits{j=1}^{2\eta}\:{\normalsize a_j}=\large\frac{1}{2\eta}\sum_\limits{j=1}^{\eta} \:{\normalsize a_j}\:\:+\:\:\large\frac{1}{2\eta}\sum_\limits{j=\eta +1}^{2\eta} \:\normalsize {a_j} \:\:\:$and,

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\:\:\:\:\large\prod_\limits{j=1}^\eta\:\sqrt[\large 2\eta] {\normalsize a_j}=\sqrt {\large\prod_\limits{j=1}^\eta \sqrt[\large \eta] {\normalsize a_j}\:\:\large\prod_\limits{j=\eta +1}^{2\eta} \sqrt[\large \eta] {\normalsize a_j}}\:\:$.

If $\:a_j\leq\large\frac{d}{da_j}$$(a_j)\:\:\forall j\:\:then,\:\:\large\frac{1}{\eta}\sum_\limits{j=1}^\eta\:\normalsize {a_j}$$\:\leq\:\large\frac{1}{\eta}\sum_\limits{j=1}^\eta\:\frac{d}{da_j}(\normalsize a_j)\:\:$and$\:\:\large\prod_\limits{j=1}^\eta\sqrt[\large \eta] {\normalsize a_j}$$\:\leq\:\large\prod_\limits{j=1}^\eta\frac{d}{da_j}(\sqrt[\large \eta]{\normalsize {a_j}})\: Therefore, by letting\:\:\large\Psi_{\eta_1}\:$$=\large\frac{1}{\eta}\sum_\limits{j=1}^{\eta}\:\normalsize a_j\:$,$\large\:\:\Phi_{\eta_1}=\prod_\limits{j=1}^\eta\sqrt[\large \eta] {\normalsize a_j} \:\:$we have . . .

$\large\Psi_{2\eta_1}=\Psi\left(\Psi_{\eta _1},\Psi_{2\eta _{\eta +1}}\right)$$\:\geq\:$$\large\Psi\left(\Phi_{\eta_1},\Phi_{2\eta_{\eta+1}}\right)$$\:\geq\:$$\large\Phi\left(\Phi_{{\eta_1}},\Phi_{2\eta_{\eta+1}}\right)=\large\Phi_{2\eta_1}$

This proves that $\:\large\Psi$$\:\geq\:$$\large\Phi\:\:$when$\:\eta\:$is a power of$\:\:2$.

$\\$

Now, if $\:\nexists j\:\:$such that $\:\eta=2^j\:$, then we let $\:\omega=\kappa^j\:\:$such that$\:\:\kappa>2,\:\:$and we let every term of the sequence $\:\large\{$$a_j\large\}$$_{ _{\large{\eta +1}}}^{ ^{\Large\omega}}=\large\Psi_{\eta_1}\:\:\:\:\normalsize \to \:\:...$

$\:\:\:\:\:\:\left(\frac{\LARGE\prod_\limits{j=1}^\eta \Large a_j}{\Large\Psi_{\eta_1}^{\omega + \eta}}\right)^{\LARGE\frac{1}{\omega}}\leq\:\large\Psi_{\omega_1} \normalsize ,\:\:\:\:$ since $\large\omega\:$is a power of $\:2\:$

Because$\:\:\large\frac{1}{\eta}\sum_\limits{j=1}^\omega\:{\normalsize a_j}=\large\Psi_{\eta_1}$, it follows that$\:\left(\large\Psi_{\omega_1}\right)^{\large\frac{1}{\omega}}\cdot \left(\large\Psi_{\eta_1}\right)^{{\large1-\frac{\eta}{\omega}}}\leq\:\left(\large\Psi_{\eta_1}\right)^{{\large-\frac{\eta}{\omega}}}$

Then, by putting both sides of the last inequality at the $\:(\large\frac{\omega}{\eta})^{th}\:$power yields :

$\:\large\prod_\limits{j=1}^\eta\sqrt[\large \eta] {\normalsize a_j}\:\leq\:\large\frac{1}{\eta}\sum_\limits{j=1}^\eta\:{\normalsize a_j}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Large\clubsuit$

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