It is a well known fact that for positive reals $x_1, x_2, \dots, x_n$, their arithmetic mean is no less than their geometric mean:
$$ \frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n} $$
and has got a multitude of proofs.
One proof approach (which I haven't seen, and what this question is about) would be to prove for the case when the numbers are positive integers. Then because of the homogenity, we can pass on to the rationals, and because rationals are dense, we move onto the reals, completing the proof.
The inequality can be rewritten as
$$ (x_1 + x_2 + \dots + x_n)^n \ge n^n x_1 x_2 \dots x_n$$
The left side can be interpreted as the number of $n$-tuples that can be formed by picking from $x_1 + x_2 + \dots + x_n$ items with replacement. So this leads to the question:
Can we give a combinatorial proof of the above (rewritten) inequality*, when the numbers involved are positive integers?
If such a proof already exists, a reference would suffice. Please feel free to add answers even if they work for specific $n \gt 1$.
*Combinatorial proofs for inequalities are not unheard of. For instance to prove that $2^n \gt n$, we count the total number of subsets of $\{1,2, \dots, n\}$ and compare with the number of subsets of size $1$.
Best Answer
I'm not completely sure this is what you're looking for, but after playing a bit with the suggested inequality, I think I have found a combinatorial way of thinking about it. In fact, we shall prove the following inequality of which the inequality we are interested in is a special case (in fact, they are equivalent):
Inequality. Let $m,n\in\mathbb{N}$ and let $k_1,k_2,\cdots k_n\in\mathbb{N}$ (yes, this is the same $n$) be natural numbers such that $k_1+k_2+\cdots+k_n=mn$. Then the following inequality holds: $$m^n \geq k_1k_2\cdots k_n\qquad(*)$$
(Taking $m=x_1+x_2+\cdots+x_n$ and $k_i = nx_i$ yields the inequality in the question.)
We shall proceed by a constructiong a sequence of injections. The following lemma is easy to see. (It is basically just the combinatorial interpretation of the inequality $k l\leq (k+1)(l-1)$ which holds for natural numbers $k<l$.)
Lemma. Let $k,l\in\mathbb{N}$ be numbers such that $k<l$. Let $A,B$ be sets such that $|A|=k$ and $|B|=l$ and let $b\in B$. Then there is an injection $\phi:A\times B\to (A\cup\lbrace b\rbrace)\times(B\setminus\lbrace b\rbrace)$.
Proof. Since $k<l$, we have $k\leq l-1$, which means that there is an injection $\psi:A\to B\setminus\lbrace b\rbrace$, so we can define $\phi$ by $$\phi(x,y)=\begin{cases}(x,y);&\textrm{if }y\neq b,\\(b,\psi(x));&\textrm{if }y=b.\end{cases}$$ It is straightforward to verify that this is indeed an injection, so we are done.
From this lemma we get the following proposition, which is the inequality $(*)$ interpreted combinatorially.
Proposition. Let $m,n\in\mathbb{N}$ and let $A_1,A_2,\cdots,A_n$ be disjoint finite sets such that $\sum_{i=1}^n|A_n|=nm$. Let $B$ be a set such that $|B|=m$. Then there is an injection $\phi:A_1\times A_2\times\cdots\times A_n\to B^n$. ($B^n$ denotes $n$-tuples of elements from $B$, as usual.)
Proof. If $|A_i|=m$ for each $i$, there is not much to prove. Otherwise, there are indices $i_1$ and $i_2$ such that $|A_{i_1}|<m$ and $|A_{i_2}|>m$. Now choose some $c\in A_{i_2}$ and define a new partition of $\bigcup_{i=1}^n A_i$ as follows: $A_i^{(1)} = A_i$ for $i\neq i_1,i\neq i_2$ and $A_{i_2}^{(1)}=A_{i_2}\setminus\lbrace c\rbrace$, $A_{i_1}^{(1)}=A_{i_1}\cup\lbrace c\rbrace$. The above lemma allows us to construct an injection $\phi_1:A_1\times A_2\times\cdots\times A_n\to A_1^{(1)}\times A_2^{(1)}\times\cdots\times A_n^{(1)}$.
If now $|A_i^{(1)}|=m$ for each $i$, we are done, otherwise, repeat the same procedure, to obtain a partition $(A_{i}^{(2)})_i$ and an injection $\phi_2:A_1^{(1)}\times A_2^{(1)}\times\cdots\times A_n^{(1)}\to A_1^{(2)}\times A_2^{(2)}\times\cdots\times A_n^{(2)}$.
The procedure will terminate after finitely many steps, since at each step $j$ the sum $\sum_{i=1}^{n}|k_i^{(j)}-m|$ decreases so it will eventually reach zero at some step $r$. (Here we define $k_i^{(j)}=|A_i^{(j)}|$.) But this implies $k_i^{(r)} = m$ for each $i$, so $|A_i^{(r)}|=|B|$ for each $i$, so there is a bijection $\psi:A_1^{(r)}\times A_2^{(r)}\times\cdots\times A_n^{(r)}\to B^n$.
This means we have found a sequence of injections: $$A_1\times A_2\times\cdots\times A_n\overset{\phi_1}{\to}A_1^{(1)}\times A_2^{(1)}\times\cdots\times A_n^{(1)}\overset{\phi_2}{\to}\cdots\overset{\phi_r}{\to}A_1^{(r)}\times A_2^{(r)}\times\cdots\times A_n^{(r)}\overset{\psi}{\to}B^n$$ and the proof is complete.
Conclusion. We have constructed a sequence of injections that proves $(*)$, thus proving the A-G inequality in a combinatorial fashion. The combinatorial meaning of the proof is basically something like this: suppose we have a set $A$ and we partition it into $n$ non-empty parts. Then the number of ways to choose a single element from each set of the partition will be the greatest when the partition is into sets of equal size.