Polynomial identity in positive terms, including AM-GM bound

Question

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.

Answer 1

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.