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} $
 A: 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.
