Can any cyclic polynomial in $a, b, c$ be expressed in terms of $a^2b+b^2c+c^2a$, $a+b+c$, $ab+bc+ca$ and $abc$?

Question

Problem. Let $f(a,b,c)$ be a cyclic polynomial in $a, b, c$. Can $f$ always be expressed as $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)$ for some polynomial $g(p, q, r, Q)$?

Motivation: If we want to prove $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)\ge 0$, and $g(p, q, r, Q)$ is non-increasing with $Q$, by using the known inequality $a^2b + b^2c + c^2a \le \frac{4}{27}(a+b+c)^3 - abc$(see How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$), sometimes, we may prove $g(a+b+c, ab+bc+ca, abc, \frac{4}{27}(a+b+c)^3 - abc)\ge 0$. As a result, we deal with a symmetric inequality rather than the original cyclic inequality.

Is it known? Is it easily to prove?

Let us see some examples. Denote $p = a+b+c, q = ab+bc+ca, r = abc$ and $Q = a^2b+b^2c+c^2a$.

1) $ab^2 + bc^2 + ca^2 = (a+b+c)(a^2+b^2+c^2) - a^3-b^3-c^3 - (a^2b+b^2c+c^2a)$

2) From the known identity $a^3b+b^3c+c^3a +(ab+bc+ca)^2 = (a+b+c)(a^2b+b^2c+c^2a+abc)$, we have $a^3b+b^3c+c^3a = p(Q + r) - q^2$.

3) $a^3b^2+b^3c^2+c^3a^2 = Qq - r(p^2-2q) - rq$.

4) $(a^2+b)(b^2+c)(c^2+a) = \cdots$

Any comments and solutions are welcome.

Answer 1

Let $R$ be the ring generated by the symmetric polynomials and $f=a^2b+b^2c+c^2a$. Certainly $f$ also contains $g=ab^2+bc^2+ca^2$ since $f+g$ is symmetric.

Let $$u_{m,n}=a^mb^n+b^mc^n+c^ma^n.$$ To prove that $R$ contains all cyclic polynomials, it suffices to prove that $u_{m,n}\in R$ for all $m$ and $n$. Indeed it suffices to prove this when $m>n\ge1$. Then, $$(a+b+c)u_{m,n}=u_{m+1,n}+u_{m,n+1}+abc u_{m-1,n-1}\tag1$$ (when $n\ge1$) and $$(ab+bc+ca)u_{m,n}=u_{m+1,n+1}+abcu_{m-1,n}+abc u_{m,n-1}\tag2$$ (when $n\ge1$).

If we proceed by induction on $m+n$, (2) shows that if all $u_{m,n}\in R$ for $r\le k$ then so do all $u_{m,n}$ with $m+n=k+1$ save perhaps $u_{k,1}$. But applying (1) for $(m,n)=(k-1,1)$ gives $$u_{k,1}=(a+b+c)u_{k-1,1}-u_{k-1,2}-abcu_{k-2,0}.$$ As we have $u_{k-2,2}\in R$, then $u_{k,1}\in R$ too.

Answer 2

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.

Thus, $$\sum_{cyc}a^2b=\frac{1}{2}\sum_{cyc}2a^2b=\frac{1}{2}\sum_{cyc}(a^2b+a^2c-(a^2c-a^2b))=$$ $$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}(a-b)(a-c)(b-c).$$ Now, for $a\geq b\geq c$ we obtain: $$\sum_{cyc}a^2b=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}\sqrt{27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)}$$ and for $a\leq b\leq c$ we obtain:

$$\sum_{cyc}a^2b=\frac{1}{2}(9uv^2-3w^3)-\frac{1}{2}\sqrt{27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)}.$$

By the similar way we can write any cyclic polynomial as function of elementary symmetric polynomials.

Because any Schur's polynomial we can write as $\prod\limits_{cyc}(a-b)P(a,b,c),$ where $P$ is a symmetric polynomial.

For example, $$\sum_{cyc}a^3b=\frac{1}{2}\sum_{cyc}(a^3b+a^3c)+\frac{1}{2}\sum_{cyc}(a^3c-a^3b)=$$ $$=\frac{1}{2}\sum_{cyc}(a^3b+a^3c)+\frac{1}{2}(a-b)(b-c)(c-a)(a+b+c).$$

Answer 3

Four years ago, I also encountered this problem. I can't prove but I can write $$f(a+b+c,\,ab+bc+ca,abc,a^2b+b^2c+c^2a) \to F(p,q,r,Q)$$ by my pqr tool run on Maple.

Similar to $a^3b+b^3c+c^3a,\,(a-b)(b-c)(c-a), \ldots$ The appropriate time (after my book is published) I will publish all the source code. Please forgive me.

Answer 4

I also can't not prove but I use my tool in Maple so I can write. $$f(a+b+c,\,ab+bc+ca,abc,a^2b+b^2c+c^2a) \to F(p,q,r,\Delta)$$