<s>[INST] 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 $abc$? 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.
 [/INST] 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.
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