<s>[INST] Polynomial identity in positive terms, including AM-GM bound Consider $n$ nonnegative numbers $x_1 \cdots x_n$. An easy  consequence of the AM-GM inequality
$$
\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n} 
$$
is a lower bound on a polynomial
$$
(x_1 + x_2 + \cdots + x_n)^n \geq n^n (x_1 x_2 \cdots x_n)
$$
which holds with equality iff $x_1 = x_2 = \cdots = x_n$.
Question 1 (existence):
Can one write the LHS polynomial as an identity which is a sum of only nonnegative terms, including the RHS? These terms can again be composites (other than the considered difference LHS - RHS itself), if it can be guaranteed that they are nonnegative.
Question 2 ($n=4$):
What's a formula for $n=4$?
Question 3 (general $n$):
Is there a principle for composing a formula for general $n$?
First solutions / remarks:
Here are ways of doing that for $n=2$:
$$
(x + y)^2 =  4 x  y + (x - y)^2
$$
and $n=3$:
$$
(x+y+z)^3 = 27 x y z + (x^3 + y^3 + z^3 - 3 x y z) + 3 (z-y)^2 x + 3 (x-z)^2 y+ 3 (y-x)^2 z
$$
where the second term (in brackets) is nonnegative again by AM-GM:  $ x^3 + y^3 + z^3 \geq  3 \sqrt[3]{x^3 y^3 z^3}  = 3 x y z $, or directly by the identity:
$x^3 + y^3 + z^3 - 3 xyz  = \frac12 (x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2) \ge 0$.
Remark:
The obvious advantage of  such a procedure would be that one could determine lower bounds of the LHS by any term on the RHS or weighted sum of terms on the RHS, with weights between 0 and 1. In particular, these lower bounds could be chosen according to prior knowledge: if all $x_i$ are known to be roughly equal, the AM-GM bound is a good one. If the $x_i$ are known to differ much, one would choose other terms on the RHS as lower bound.
 [/INST] Yes, we can. 
We can do it by Bricks Throwing Method.
It's very ugly but it works.
For four variables we need to work with $(a+b+c+d)^4-256abcd$.
$3(a^4+b^4+c^4+d^4)\geq\sum\limits_{cyc}a^3(b+c+d)$ because we throw one brick:
$\sum\limits_{cyc}(3a^4-a^3(b+c+d))=\frac{1}{2}\sum\limits_{sym}(a^4-a^3b-ab^3+b^4)=\frac{1}{2}\sum\limits_{sym}(a-b)^2(a^2+ab+b^2)$.
More brick trowing:
$\sum\limits_{cyc}a^3(b+c+d)\geq\frac{1}{2}\sum\limits_{sym}a^2b^2$ gives
$\sum\limits_{cyc}a^3(b+c+d)-\frac{1}{2}\sum\limits_{sym}a^2b^2=\frac{1}{2}\sum\limits_{sym}(a^3b-2a^2b^2+ab^3)=\frac{1}{2}\sum\limits_{sym}ab(a-b)^2$ and so on...
I'll write a full proof.
More brick trowing:
$\frac{1}{2}\sum\limits_{sym}a^2b^2-\sum\limits_{cyc}a^2(bc+bd+cd)=\frac{1}{2}\sum\limits_{sym}(a-b)^2(c^2+d^2)$
The last brick trowing:
$\sum\limits_{cyc}a^2(bc+bd+cd)-12abcd=\frac{1}{4}\sum\limits_{sym}(a-b)^2cd$.
After using these bricks throwing we'll write $(a+b+c+d)^4-256abcd$ like a sum of squares. 
</s>