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.
Best Answer
Power mean inequality can be used to get bounds the difference between various means. Not exactly using variances, but you have $$AM - GM \le \max_i x_i - \min_i x_i$$ Or if that's too large a bound, note $$AM - GM \le AM - HM$$
If you want variances involved, try manipulating $$\sqrt{\frac1n \sum x_i^2} - GM \ge AM-GM$$