Representing a symmetric polynomial as a conical sum of squares

Question

This question in inspired by the recent solution to another question.

The following inequality for monomial symmetric polynomials in 4 positive variables $x_1,x_2,x_3,x_4$: $$m_{(4, 3, 2, 1)} + m_{(4, 4, 2)} \geq 3m_{(4, 2, 2, 2)} + m_{(4, 3, 3)} + 2 m_{(4, 4, 1, 1)}$$ seems to be hard to prove directly, but easily follows from a rather unobvious identity: \begin{split} &m_{(4, 3, 2, 1)} + m_{(4, 4, 2)} - (3m_{(4, 2, 2, 2)} + m_{(4, 3, 3)} + 2 m_{(4, 4, 1, 1)})\\ =&\frac12\left( f(x_1,x_2,x_3,x_4)^2 + f(x_1,x_3,x_2,x_4)^2 + f(x_2,x_3,x_1,x_4)^2 \right), \end{split} with $$f(x,y,z,t) := (x-y) (z-t) (xy(z+t) - (x+y)zt).$$

Relatedly, my questions are

Q1: How to find representation of a given symmetric polynomial in a fixed number of variables as a conical sum of squares of polynomials if one exists?

Q2: Is it any simpler if we restrict Q1 to the case when all the coefficients in the sum are the same, and the polynomials being squared represent the same polynomial with just somehow permuted arguments? (like in the above example)

Q3: Is there a direct "obvious" proof of the above inequality.

Answer 1

For four variables $a$, $b$ $c$ and $d$ we'll use the following natation. $$\sum_{sym}a=6(a+b+c+d),$$ $$\sum_{sym}ab=4(ab+ac+bc+ad+bd+cd),$$ $$\sum_{sym}a^2b=2\sum_{cyc}a^2(b+c+d).$$ Number of addends is equal to $4!=24$.

Now, let $x_1=\frac{1}{a},$ $x_2=\frac{1}{b},$ $x_3=\frac{1}{c}$ and $x_4=\frac{1}{d}.$

Thus, we need to prove that: $$\sum_{sym}\left(\frac{2}{a^4b^3c^2d}+\frac{1}{a^4b^4c^2}-\frac{1}{a^4b^2c^2d^2}-\frac{1}{a^4b^3c^3}-\frac{1}{a^4b^4cd}\right)\geq0$$ or $$\sum_{sym}(2a^3b^2c+a^4b^2-a^2b^2c^2-a^4bc-a^3b^3)\geq0$$ which is $$\sum_{sym}(a-b)^2(a-c)^2(b-c)^2\geq0$$ because for three variables $a$, $b$ and $c$ we have: $$\prod_{cyc}(a-b)^2=\sum_{cyc}(a^4b^2+a^4c^2-2a^3b^3-2a^4bc+2a^3b^2c+2a^3c^2b-2a^2b^2c^2)=$$ $$=\sum_{sym}(a^4b^2-a^4bc-a^3b^3+2a^3b^2c-a^2b^2c^2).$$ The last identity we can get by the following way: $$\sum_{cyc}(a^4b^2+a^4c^2-2a^4bc-2a^3b^3+2a^3b^2c+2a^3c^2b-2a^2b^2c^2)=$$ $$=\sum_{cyc}(c^4a^2-2c^4ab+c^4b^2-c^3(a^3+b^3-a^2b-ab^2)+abc(a^2c+b^2c-2abc))=$$ $$=\sum_{cyc}(a-b)^2(c^4-c^3(a+b)+c^2ab)=\sum_{cyc}(a-b)^2c^2(c-a)(c-b)=$$ $$=\sum_{cyc}(a-b)(b-c)(c-a)c^2(b-a)=\prod_{cyc}(a-b)\sum_{cyc}(a^2c-a^2b)=\prod_{cyc}(a-b)^2.$$ Your linked inequality:

For positive variables $x_1$, $x_2$, $x_3$ and $x_4$ we have $$24m_{(3, 3, 3, 1)} + 3 m_{(4, 3, 2, 1)} + 9 m_{(4, 4, 2)} \geq 12 m_{(3, 3, 2, 2)} + 3 m_{(4, 2, 2, 2)} + 9 m_{(4, 3, 3)} + 14 m_{(4, 4, 1, 1)}$$

We can prove by the following way.

Again, let $x_1=\frac{1}{a},$ $x_2=\frac{1}{b},$ $x_3=\frac{1}{c}$ and $x_4=\frac{1}{d}.$

Thus, in the previous notation we need to prove that: $$\sum_{sym}(9a^4b^2-9a^4bc+8a^4bcd+6a^3b^2c-6a^2b^2cd-7a^3b^3-a^2b^2c^2)\geq0.$$ Now, let $$a+b+c+d=4u,$$ $$ab+ac+bc+ad+bd+cd=6v^2,$$$$abc+abd+acd+bcd=4w^3$$ and $$abcd=t^4.$$ Thus, we need to prove that: $$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0.$$ Now, $a$, $b$, $c$ and $d$ are roots of the polynomial $$f(p)=(p-a)(p-b)(p-c)(p-d)$$ or $$p^4-4up^3+6v^2p^2-4w^3p+t^4,$$ which by the Rolle's theorem says that the polynomial $$f'(p)=4p^3-12up^2+12v^2p-4w^3$$ has three roots.

Let $x$, $y$ and $z$ be these roots.

Thus, $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$, which says that it's enough to prove our inequality for three variables $x$, $y$ and $z$.

But it's really incredible here that: $$(x-y)^2(x-z)^2(y-z)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ and we are done!