Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have $$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$

Very nice. Thanks.

MariusStanean wrote: Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have $$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$ what is Cebasev inequality

zsgvivo wrote: MariusStanean wrote: Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have $$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$ what is Cebasev inequality https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality

sqing wrote: zsgvivo wrote: MariusStanean wrote: Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have $$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$ what is Cebasev inequality https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality thanks!

@fair enogh...thanks

For $a+b<1, a,b\ge 0$ holds $\frac{1}{1-a}\le\frac{1}{1-b}\iff a\le b$ and $a(1-a)\le b(1-b)\iff a-b\le (a-b)(a+b)\iff (1-a-b)(a-b)\le0\iff a\le b$ which proves sequences $\left(\frac{1}{1-x_i}\right)$, $\left(x_i(1-x_i)\right)$ are non-decreasing for $0<x_1\le x_2\le\ldots \le x_n, x_1+x_2+...+x_n=1$ so we can use Chebyshev