Why this inequality is correct

For $sumlimits_{i=1}^mx_igeq1$ 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: $$(b-m)left(c+1-frac{c}{a}right)=sum_{i=1}^mfrac{1-x_i}{x_i}left(sum_{i=1}^mfrac{x_i}{1-x_i}left(1-frac{1}{sumlimits_{i=1}^mx_i}right)+1right)geq$$ $$geq m^2-frac{m^2}{sumlimits_{i=1}^mx_i}+sum_{cyc}frac{1-x_i}{x_i}=$$ $$=m^2-m+frac{1}{sumlimits_{i=1}^mx_i}left(sum_{i=1}^mx_isum_{i=1}^mfrac{1-x_i}{x_i}-msum_{i=1}^m(1-x_i)right)geq m^2-m.$$

The proof without condition $sumlimits_{i=1}^mx_igeq1$.

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: $$(b-m)left(c+1-frac{c}{a}right)=sum_{i=1}^mfrac{1-x_i}{x_i}left(sum_{i=1}^mfrac{x_i}{1-x_i}+1-frac{sumlimits_{i=1}^mfrac{x_i}{1-x_i}}{sumlimits_{i=1}^mx_i}right)=$$ $$=sum_{i=1}^mfrac{1-x_i}{x_i}left(sum_{i=1}^mfrac{x_i}{1-x_i}+frac{sumlimits_{i=1}^mleft(x_i-frac{x_i}{1-x_i}right)}{sumlimits_{i=1}^mx_i}right)=$$ $$=sum_{i=1}^mfrac{1-x_i}{x_i}left(sum_{i=1}^mfrac{x_i}{1-x_i}-frac{sumlimits_{i=1}^mfrac{x_i^2}{1-x_i}}{sumlimits_{i=1}^mx_i}right)=frac{sumlimits_{i=1}^mfrac{1-x_i}{x_i}sumlimits_{i=1}^mfrac{x_i(a-x_i)}{1-x_i}}{sumlimits_{i=1}^mx_i}geq$$ $$geqfrac{msumlimits_{i=1}^mleft(frac{1-x_i}{x_i}cdotfrac{x_i(a-x_i)}{1-x_i}right)}{sumlimits_{i=1}^mx_i}=frac{msumlimits_{i=1}^m(a-x_i)}{sumlimits_{i=1}^mx_i}=m(m-1).$$

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

Therefore begin{align} b-m&= mg ,\ frac{c}{a}&lefrac{(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 mgleft(frac{m-1}{g}right)\ &=m(m-1) . end{align}