<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 $a+b+c=3u$, $ab+ac+bc=3v^2$, 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, $$\sum_{cyc}a^3b=\frac{1}{2}\sum_{cyc}(a^3b+a^3c)+\frac{1}{2}\sum_{cyc}(a^3c-a^3b)=$$
$$=\frac{1}{2}\sum_{cyc}(a^3b+a^3c)+\frac{1}{2}(a-b)(b-c)(c-a)(a+b+c).$$
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