i) is easy, ii) i still dont know how to deal with x - p/n. here is part i Let t = $\sum x_i^2$. i) Rewrite as $(n-1)p^2 \geq 2qn$ $\Leftrightarrow (n-1)t \geq 2q$ But $t \geq q$ and $(n-1) \geq 2$ with equality when $n=3$ and $x_1 = x_2 = x_3$.

orl wrote: Let $p$ and $q$ be their symmetric sum of degree $1$ and $2$ respectively. What are those?

N.T.TUAN wrote: orl wrote: Let $p$ and $q$ be their symmetric sum of degree $1$ and $2$ respectively. What are those? $\sum x_i, \sum x_ix_j$

warut_suk wrote: $\sum x_ix_j$ $n!$ element in this sum? Thank you very much.

Proof:(1)By AM-QM \[ n\sum_{i=1}^nx_i^2\geq(\sum_{i=1}^nx_i)^2 \] And \[ \sum_{i=1}^nx_i^2=p^2-2q,\sum_{i=1}^nx_i=p \] So $n(p^2-2q)\geq p^2,p^2\cdot\frac{n-1}{n}-2q\geq0$. (2)We only need to prove the inequality for $x_n$. \[ \sum_{i=1}^{n-1}x_i=p-x_n,\sum_{i=1}^{n-1}x_i^2=p^2-2q-x_n^2 \] By AM-QM \[ (n-1)\sum_{i=1}^{n-1}x_i^2\geq(\sum_{i=1}^{n-1}x_i)^2 \] So $(n-1)(p^2-2q-x_n^2)\geq(p-x_n)^2$, $nx_n^2-2px_n+2(n-1)q-(n-2)p^2\leq0$ By (1)\[ (x_n-\frac{p}{n})^2\leq\frac{p^2}{n^2}+\frac{n-2}{n}p^2-\frac{2(n-1)}{n}q=\frac{(n-1)^2}{n^2}p^2-2\frac{n-1}{n}q\geq0 \] So \[ |x_n-\frac{p}{n}|\leq\frac{n-1}{n}\sqrt{p^2-\frac{2nq}{n-1} }\] And we can also prove it for $x_1,x_2,\ldots,x_{n-1}$ by the same discussion. So \[ |x_i-\frac{p}{n}|\leq\frac{n-1}{n}\sqrt{p^2-\frac{2nq}{n-1} }\] for every meaningful $i$.

ii) can also be proven if we look at it as quadratic inequality towards $x_i$ and show ( using i) ) that discriminant is negative iff $(n-1)p^2 - 2nq>0$ and if the last is $0$ then all $x_i$ are equal and LHS of ii) is 0

ii) can also be proven if we look at it as quadratic inequality towards $x_i$ and show ( using i) ) that discriminant is negative iff $(n-1)p^2 - 2nq>0$ and if the last is $0$ then all $x_i$ are equal and LHS of ii) is 0