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I meet this following problem If $$n\ge 3,\sum_{i=1}^{n}\left(\prod_{j\neq i}(a_{i}-a_{j})\right)\ge 0$$ where $a_{i}$ are real numbers.

when $n=3$, it is Schur's inequality so which $n$ such this inequality?

but for more generalization form of Schur'.s Inequality exists?

  • For $n=3$, that looks like $a_1^2+a_2^2+a_3^2 \ge a_1 a_2 + a_2 a_3 + a_3 a_1$, which is as much Schur as it is rearrangement or AM-GM. – Macavity Jul 02 '15 at 09:29
  • Further, for $n=4$, check $a_i = (9, 8, 4, 2)$ to get a contradiction, so that inequality isn't true even for positives. – Macavity Jul 02 '15 at 09:36
  • Thank you, but $n\ge 5$ is also isn't true? –  Jul 02 '15 at 09:52
  • I haven't checked, but unless you have a strong reason to believe it will be, it may be a lot of ultimately fruitless effort. There are other ways in which Schur is generalised - though really not found anything as elegant as the 3 variable case. – Macavity Jul 02 '15 at 09:54
  • I have find this paper: http://www.jwss.cc/?q=A+Schur+Type+Inequality+for+six++Variables+ –  Jul 02 '15 at 10:04

2 Answers2

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Now I have solve it, this inequality it is only true for $n=3,n=5$ holds.

For $n=3$ then inequality can write as $$(a_{1}-a_{2})(a_{1}-a_{3})+(a_{2}-a_{1})(a_{2}-a_{3})+(a_{3}-a_{1})(a_{3}-a_{2})=a^2_{1}+a^2_{2}+a^2_{3}-a_{1}a_{2}-a_{1}a_{3}-a_{2}a_{3}\ge 0$$ it is clear true

for $n=5$,WLOG, we assume that $$a_{1}\ge a_{2}\ge a_{3}\ge a_{4}\ge a_{5}$$ Lemma 1: $$I=(a_{1}-a_{2})(a_{1}-a_{3})(a_{1}-a_{4})(a_{1}-a_{5})+(a_{2}-a_{1})(a_{2}-a_{3})(a_{2}-a_{4})(a_{2}-a_{5})\ge 0$$ Proof: because $$I=(a_{1}-a_{2})[(a_{1}-a_{3})(a_{1}-a_{4})(a_{1}-a_{5})-(a_{2}-a_{3})(a_{2}-a_{4})(a_{2}-a_{5})]$$ since $$a_{1}-a_{3}\ge a_{2}-a_{3}\ge 0$$ $$a_{1}-a_{4}\ge a_{2}-a_{4}\ge 0$$ $$a_{1}-a_{5}\ge a_{2}-a_{5}\ge 0$$ $$a_{1}-a_{2}\ge 0$$ so $$I\ge 0$$

Use same methods we have $$(a_{5}-a_{1})(a_{5}-a_{2})(a_{5}-a_{3})(a_{5}-a_{4})+(a_{4}-a_{1})(a_{4}-a_{2})(a_{4}-a_{3})(a_{4}-a_{5})\ge 0$$ and $$(a_{3}-a_{1})(a_{3}-a_{2})(a_{3}-a_{4})(a_{3}-a_{5})\ge 0$$ so for $n=5$ is hold

for $n=4$, you can take $a_{1}=0,a_{2}=a_{3}=a_{4}=1$

for $n\ge 6$,you can take $$a_{1}=0,a_{2}=a_{3}=a_{4}=1,a_{5}=a_{6}=\cdots=a_{n}=-1$$ then $$\sum_{i=1}^{n}\left(\prod_{j\neq i}(a_{i}-a_{j})\right)=(-1)^3<0$$

math110
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Here are some inequalities:

  • Schur's inequality: If $a,b,c,p \geq 0$, then $$a^p(a-b)(a-c)+b^p(b-c)(b-a)+c^p(c-b)(c-a) \geq 0$$
  • Weak 6-variable Schur: Suppose $a \geq b \geq c$, $x+z \geq y \geq 0$. Then $$x(a-b)(a-c)+y(b-c)(b-a)+z(c-b)(c-a) \geq 0$$
  • Strong 6-variable Schur: Suppose $a \geq b \geq c$, $x+z \geq y \geq 0$. Then $$x^2(a-b)(a-c)+y^2(b-c)(b-a)+z^2(c-b)(c-a) \geq 0$$

Now we get to some inequalities with functions inside it. $f: \mathbb R \rightarrow \mathbb R^+$ be a function that can be expressed as the sum of monotonic non-negative functions. Let $g: \mathbb R \rightarrow \mathbb R$ and $h: \mathbb R \rightarrow \mathbb R$ be odd (functions such that $f(x)=-f(-x)$) and increasing functions.

  • Vornicu-Schur inequality: Let $a,b,c,x,y,z \in \mathbb R$, $a \geq b \geq c$ and either $x \geq y \geq z$ or $z \geq y \geq x$. Let $k$ be a postive integer. Then $$f(x)(a-b)^k(a-c)^k+f(y)(b-a)^k(b-c)^k+f(z)(c-a)^k(c-b)^k \geq 0 $$

We get Schur's inequality if $k=1$, $x=a$, $y=b$, $x=c$, $f(n)=n^p$. But we can generalize further.

  • Weak generalized Schur inequality: Let $f,g,h$ be defined as above and let $a,b,c \in \mathbb R$, $a \geq b \geq c$. Then $$f(a) g(h(a-b)h(a-c))+ f(b) g(h(b-a)h(b-c))+ f(c) g(h(c-a)h(c-b)) \geq 0$$
  • Strong generalized Schur inequality: Let $f,g,h$ be defined as above and let $a,b,c \in \mathbb R$, $a \geq b \geq c$. We further require $f$ to be convex. Then $$f(a)^2 g(h(a-b)h(a-c))+ f(b)^2 g(h(b-a)h(b-c))+ f(c)^2 g(h(c-a)h(c-b)) \geq 0$$
wythagoras
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