SOS: Proof of the AM-GM inequality

Yes, we can write $x_1^n+x_2^n+...+x_n^n-nx_1x_2...x_n$ in the SOS form, but I think it's very ugly and it's not useful for big $n$.

We can make the following.

For $n=3$: $$a^3+b^3+c^3-3abc=\sum_{cyc}\left(a^3-\frac{1}{2}a^2b-\frac{1}{2}a^2c\right)+\sum_{cyc}\left(\frac{1}{2}a^2b+\frac{1}{2}a^2c-abc\right)=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b)+\frac{1}{2}\sum_{cyc}c(a-b)^2=\frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2.$$

For $n=4$: $$a^4+b^4+c^4+d^4-4abcd=\sum_{cyc}\left(a^4-\frac{1}{3}(a^3b+a^3c+a^3d)\right)+$$ $$+\frac{1}{6}\sum_{sym}(a^3b-a^2b^2)+\frac{1}{6}\sum_{sym}(a^2b^2-a^2bc)+\frac{1}{3}\sum_{cyc}(a^2bc+a^2cd+a^2bd-3abcd)=$$ $$=\frac{1}{6}\left(\sum_{sym}(a^4-a^3b)+\sum_{sym}(a^3b-a^2b^2)+\sum_{sym}(a^2b^2-a^2bc)+\sum_{sym}(a^2bc-abcd)\right)=$$ $$=\frac{1}{12}\left(\sum_{sym}(a^4-a^3b-ab^3+b^4)+\sum_{sym}(a^3b-2a^2b^2+ab^3)+\sum_{sym}(c^2a^2-2c^2ab+c^2b^2)+\sum_{sym}(a^2cd-2abcd+b^2cd)\right)=$$ $$=\frac{1}{12}\sum_{sym}(a-b)^2(a^2+b^2+c^2+2ab+cd)=$$ $$=\tfrac{(a-b)^2(2(a+b)^2+(c+d)^2)+(a-c)^2(2(a+c)^2+(b+d)^2)+(a-d)^2(2(a+d)^2+(b+c)^2)}{6}+$$ $$+\tfrac{(b-c)^2(2(b+c)^2+(a+d)^2)+(b-d)^2(2(b+d)^2+(a+c)^2)+(c-d)^2(2(c+d)^2+(a+b)^2)}{6}.$$ Here $$\sum\limits_{sym}a=6(a+b+c+d)$$ $$\sum\limits_{sym}a^2b^2=4(a^2b^2+a^2c^2+b^2c^2+a^2d^2+b^2d^2+c^2d^2),...$$ For $n=5$: $$a^5+b^5+c^5+d^5+e^5-5abcde=\frac{1}{24}\sum_{sym}(a^5-abcde)=$$ $$=\frac{1}{24}\left(\sum_{sym}(a^5-a^4b)+\sum_{sym}(a^4b-a^3b^2)+\sum_{sym}(a^3b^2-a^3bc)\right)+$$ $$+\frac{1}{24}\left(\sum_{cyc}(a^3bc-a^2b^2c)+\sum_{sym}(a^2b^2c-a^2bcd)+\sum_{sym}(a^2bcd-abcde)\right)=$$ $$=\frac{1}{48}\left(\sum_{sym}(a^5-a^4b-ab^4+b^5)+\sum_{sym}(a^4b-a^3b^2-a^2b^3+ab^4)+\sum_{sym}(c^3a^2-2c^3ab+c^3b^2)\right)+$$ $$+\frac{1}{48}\left(\sum_{cyc}(a^3bc-2a^2b^2c+b^3ac)+\sum_{sym}(a^2c^2d-2c^2abd+b^2c^2d)+\sum_{sym}(a^2cde-2abcde+b^2cde)\right)=$$ $$=\frac{1}{48}\sum_{sym}(a-b)^2(a^3+2a^2b+2ab^2+b^3+c^3+abc+c^2d+cde).$$ The rest is the same.