Why this inequality is correct

Question

Let $0<x_1\leq\dots\leq x_m<1$, I denote $$a=\sum_{i=1}^mx_i,\qquad b=\sum_{i=1}^m\frac{1}{x_i},\qquad c=\sum_{i=1}^m\frac{x_i}{1-x_i}.$$ Im trying to prove that $$(b-m)(c+1-\frac{c}{a})\geq m(m-1).$$ I did prove it for the case of $a\geq 1$ using the Cauchy-Schwartz inequality. I implement a simulation in Python to construct a counterexample but in vain.

Answer 1

For $\sum\limits_{i=1}^mx_i\geq1$ it's nice:

Since $(x_1,x_2,...,x_m)$ and $\left(\frac{1-x_1}{x_1},\frac{1-x_2}{x_2},...,\frac{1-x_m}{x_m}\right)$ have opposite ordering,

by C-S and Chebyshov we obtain: $$(b-m)\left(c+1-\frac{c}{a}\right)=\sum_{i=1}^m\frac{1-x_i}{x_i}\left(\sum_{i=1}^m\frac{x_i}{1-x_i}\left(1-\frac{1}{\sum\limits_{i=1}^mx_i}\right)+1\right)\geq$$ $$\geq m^2-\frac{m^2}{\sum\limits_{i=1}^mx_i}+\sum_{cyc}\frac{1-x_i}{x_i}=$$ $$=m^2-m+\frac{1}{\sum\limits_{i=1}^mx_i}\left(\sum_{i=1}^mx_i\sum_{i=1}^m\frac{1-x_i}{x_i}-m\sum_{i=1}^m(1-x_i)\right)\geq m^2-m.$$

The proof without condition $\sum\limits_{i=1}^mx_i\geq1$.

Since $$\left(\frac{1-x_1}{x_1},\frac{1-x_2}{x_2},...,\frac{1-x_m}{x_m}\right)$$ and $$\left(\frac{x_1(a-x_1)}{1-x_1},\frac{x_2(a-x_2)}{1-x_2},...,\frac{x_m(a-x_m)}{1-x_m}\right)$$ have an opposite ordering, by Chebyshov we obtain: $$(b-m)\left(c+1-\frac{c}{a}\right)=\sum_{i=1}^m\frac{1-x_i}{x_i}\left(\sum_{i=1}^m\frac{x_i}{1-x_i}+1-\frac{\sum\limits_{i=1}^m\frac{x_i}{1-x_i}}{\sum\limits_{i=1}^mx_i}\right)=$$ $$=\sum_{i=1}^m\frac{1-x_i}{x_i}\left(\sum_{i=1}^m\frac{x_i}{1-x_i}+\frac{\sum\limits_{i=1}^m\left(x_i-\frac{x_i}{1-x_i}\right)}{\sum\limits_{i=1}^mx_i}\right)=$$ $$=\sum_{i=1}^m\frac{1-x_i}{x_i}\left(\sum_{i=1}^m\frac{x_i}{1-x_i}-\frac{\sum\limits_{i=1}^m\frac{x_i^2}{1-x_i}}{\sum\limits_{i=1}^mx_i}\right)=\frac{\sum\limits_{i=1}^m\frac{1-x_i}{x_i}\sum\limits_{i=1}^m\frac{x_i(a-x_i)}{1-x_i}}{\sum\limits_{i=1}^mx_i}\geq$$ $$\geq\frac{m\sum\limits_{i=1}^m\left(\frac{1-x_i}{x_i}\cdot\frac{x_i(a-x_i)}{1-x_i}\right)}{\sum\limits_{i=1}^mx_i}=\frac{m\sum\limits_{i=1}^m(a-x_i)}{\sum\limits_{i=1}^mx_i}=m(m-1).$$

Answer 2

Let $\ \displaystyle g=\frac{b}{m}-1=\frac{1}{m}\sum_{i=1}^m\frac{1-x_i}{x_i}\ $, the arithmetic mean of $\ \displaystyle \frac{x_1}{1-x_1},$$\ \displaystyle \frac{x_2}{1-x_2},$$\dots$$\displaystyle \frac{x_m}{1-x_m}\ $. Since $\ \displaystyle\frac{m}{c}\ $ is their harmonic mean, $\ \displaystyle g\ge\frac{m}{c}\ $ by the harmonic mean-arithmetic mean inequality. Likewise, $\ \displaystyle\frac{1}{1+g}=\frac{b}{m}\ $ is the harmonic mean of $\ x_1,$$x_2,$$\dots,$$x_m\ $, and $\ \displaystyle\frac{a}{m}\ $ is their arithmetic mean, so $\ \displaystyle\frac{1}{1+g}\le$$\displaystyle\frac{a}{m}\ $.

Therefore \begin{align} b-m&= mg\ ,\\ \frac{c}{a}&\le\frac{(1+g)c}{m}\ ,\\ c+1-\frac{c}{a}&\ge c+1-\frac{(1+g)c}{m}\ ,\\ &=\frac{c(m-1-g)}{m}+1\\ &\ge \frac{(m-1-g)}{g}+1\\ &= \frac{m-1}{g}\ \text{, and}\\ (b-m)\left(c+1-\frac{c}{a}\right)&\ge mg\left(\frac{m-1}{g}\right)\\ &=m(m-1)\ . \end{align}