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(Background) Let $S_a$, $S_b$, $S_c$ be functions of $a$, $b$, $c$. Then the function \[f(a,b,c)=S_a(b-c)^2+S_b(c-a)^2+S_c(a-b)^2\tag1\] can be proven to be $\ge0$ under either one of a few conditions. This is given by the $\sf Phan~Kim~Hung$ theorem. It made me think of another function with similar structure, that is \[S_a(a-b)(a-c)+S_b(b-a)(b-c)+S_c(c-a)(c-b).\tag2\] In some inequalities, I sometimes find it easier to express in $(2)$ rather than $(1)$ (or any other form of "sum of squares"). So are there any conditions that can prove that $(2)\ge0$?

For example, $S_a=a^t$ satisfies, because it is $\sf Schur$ inequality. What about more general ones?

Admittedly, such conditions, even if it exist, it will be much stricter because the $(a-b)(b-c)$ parts may not even be positive. But I'd still like to know if there're any known theories about this.

  • Well, there could be many conditions under which this holds true - for e.g. If $S$ is any increasing function of its argument it works. Hence may be difficult to answer your question in a unique or "best" way. – Macavity Jan 28 '23 at 14:32
  • See also: https://math.stackexchange.com/questions/1346798/do-more-generalizations-of-schurs-inequality-exist – Macavity Jan 28 '23 at 14:38
  • @Macavity Just take any condition that makes it seem practical will be good. –  Jan 28 '23 at 14:44
  • Well, in that you already have many in the comments above. – Macavity Jan 28 '23 at 15:22

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Since $$\frac{1}{2}\sum_{cyc}\left(S_a+S_b-S_c\right)(a-b)^2=\frac{1}{2}\sum_{cyc}(S_a+S_b-S_c)(a-b)(a-c+c-b)=$$ $$=\frac{1}{2}\sum_{cyc}(S_a+S_b-S_c+S_c+S_a-S_b)(a-b)(a-c)=\sum_{cyc}S_a(a-b)(a-c),$$ we can use the Phan Kim Hung's conditions.

Michael Rozenberg
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