# My proof of the inequality of arithmetic and geometric means

Doing the exercises from the Apostol's Calculus I give my proof that the geometric mean is less than or equal the arithmetic mean ($G \le M_1$). I followed the hints from the book and I think I've given the very proof the author had in mind, but not sure that it is correct. If it is and you've never seen it than it can enhance your knowledge. Because I think that this proof is not present at the wikipedia page about this inequality, which offers rather complicated proofs. But this proof is elementary and rigorous (if it's correct).

1) Let first proove this lemma.

Lemma: If we have $n$ numbers and their product is equal to $1$, then their sum is greater than or equal to $n$. ($a_1 a_2\cdots a_n = 1 \implies a_1 + a_2 + \ldots + a_n \ge n$).

Proof:

Let's $a_1a_2\cdots a_n = 1$. We want to determine $a_1 + a_2 + \ldots + a_n$.

If all the numbers are equal to $1$ that their sum is equal to $n$. Otherwise some numbers are less than $1$ and some are greater than $1$.

Let $a_1 < 1$ and $a_2 > 1$. Then $(1 - a_1)(a_2 - 1) > 0 \implies a_1 + a_2 > 1 + a_1a_2.$

$a_1 + a_2 + a_3 + \cdots + a_n > 1 + a_1a_2 + a_3 + \cdots + a_n$

$b = a_1a_2$

$ba_3\cdots a_n = 1$

Recursively repeting this step we find that $a_1 + a_2 + a_3 + \cdots + a_n \ge n$

2) Now we can easily deduce the inequality $G \le M_1$ from the lemma.

Let $a_1, a_2, \ldots, a_n$ be arbitrary positive real numbers. There is some number $k$ such that $a_1a_2\cdots a_nk^n = 1$

$(a_1k)(a_2k)\cdots (a_nk) = 1$

By the lemma we obtain

$(a_1k)(a_2k)\cdots (a_nk) = 1 \implies (a_1k) + (a_2k) + \ldots + (a_nk) \ge n$

$\frac{(a_1k) + (a_2k) + \ldots + (a_nk)}{n} \ge 1$

$\frac{(a_1k) + (a_2k) + \ldots + (a_nk)}{n} \ge 1 = \sqrt[n]{(a_1k)(a_2k)\cdots (a_nk)}$

$\frac{k(a_1 + a_2 + \ldots + a_n)}{n} \ge \sqrt[n]{k^na_1a_2\cdots a_n}$

$\frac{k(a_1 + a_2 + \ldots + a_n)}{n} \ge k\sqrt[n]{a_1a_2\cdots a_n}$

$\frac{a_1 + a_2 + \ldots + a_n}{n} \ge \sqrt[n]{a_1a_2\cdots a_n}$

• The first part of the proof looks fishy, and I really don't understand how the definition being recursive (?) makes the cut. I think that either in this case, as with the usual proof, you need mathematical induction here...and in the second part you need $\;a_i\ge 0\;$ – DonAntonio Jan 17 '14 at 14:46
• Very nice. However, you should define the number $\operatorname{Sum}(n)$ to be the least lower bound of the possible values of the sum of $n$ numbers with product 1, not the value, since this will depend on the numbers. In fact, this is how you use it. This proof can be formulated more cleanly without this notation simply by using the induction hypothesis that if $a_1 \cdots a_{n-1} = 1$, then $a_1 + \cdots + a_{n-1} \geq n-1$. Finally, you meant "there is some number $k$", not "some number $k^n$." – user121926 Jan 17 '14 at 14:49
• Sorry, I meant the greatest lower bound. – user121926 Jan 17 '14 at 15:03
• Framing it as induction would help, the recursion misleads. – Macavity Jan 17 '14 at 15:21
• It's "prove this lemma", not "proof". To make it clear, you should state that $(a_2a_3,a_4,...)$ has one less number but still has product $1$, and hence you can recursively or inductively repeat the process which always decreases the expression. To make it cleaner, don't split into cases but instead note that there must be at least one number that is at least $1$ and one that is at most $1$ and just repeat the process until 1 number is left. – user21820 Jan 17 '14 at 15:38

To tackle AM-GM note that if you set $A=(a_1a_2\cdots a_n)^{1/n}$, the numbers $A_n=a_n/A$ are such that $A_1\cdots A_n=1$. Observe your $k$ is precisely $A$. This means $$\frac{a_1}A+\cdots+\frac{a_n}A\geqslant n$$ that is $$\frac{a_1+\cdots+a_n}n\geqslant (a_1\cdots a_n)^{1/n}$$ as desired.
Your proof is fine, you just need to be clear about the inductive step. Your lemma is trivial for $n=1$, so assume $n>1$, and the claim proven for $n-1$ numbers. Now, as you said, there is nothing to prove if all $a_1,\ldots,a_n=1$, so we may assume they are not all equal to $1$, which means there must exist $a_j<1$ and $a_i>1$. We may assume after reordering that $j=n$ and $i=1$. As in the proof of the case $n=1$, we thus have $a_1<1<a_n$ and thus $a_1+a_n>1+a_na_1$. Let $y=a_na_1$. Then $a_2,\ldots,a_{n-1},y$ are $n-1$ numbers whose product is $=1$, hence by our inductive hypothesis, $y+a_2+\cdots+a_{n-1}\geqslant n-1$ which means $$1+y+a_2+\cdots+a_{n-1}\geqslant n$$ but by the above $$a_1+a_n+(a_2+\cdots+a_{n-1})>(1+y)+(a_2+\cdots+a_{n-1})\geqslant n$$ so the inductive step is complete and the lemma is proven.