Why this inequality is correct

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: $$(b-m)\left(c+1-\frac{c}{a}\right)=\sum_{i=1}^m\frac{1-x_i}{x_i}\left(\sum_{i=1}^m\frac{x_i}{1-x_i}+1-\frac{\sum\limits_{i=1}^m\frac{x_i}{1-x_i}}{\sum\limits_{i=1}^mx_i}\right)=$$ $$=\sum_{i=1}^m\frac{1-x_i}{x_i}\left(\sum_{i=1}^m\frac{x_i}{1-x_i}+\frac{\sum\limits_{i=1}^m\left(x_i-\frac{x_i}{1-x_i}\right)}{\sum\limits_{i=1}^mx_i}\right)=$$ $$=\sum_{i=1}^m\frac{1-x_i}{x_i}\left(\sum_{i=1}^m\frac{x_i}{1-x_i}-\frac{\sum\limits_{i=1}^m\frac{x_i^2}{1-x_i}}{\sum\limits_{i=1}^mx_i}\right)=\frac{\sum\limits_{i=1}^m\frac{1-x_i}{x_i}\sum\limits_{i=1}^m\frac{x_i(a-x_i)}{1-x_i}}{\sum\limits_{i=1}^mx_i}\geq$$ $$\geq\frac{m\sum\limits_{i=1}^m\left(\frac{1-x_i}{x_i}\cdot\frac{x_i(a-x_i)}{1-x_i}\right)}{\sum\limits_{i=1}^mx_i}=\frac{m\sum\limits_{i=1}^m(a-x_i)}{\sum\limits_{i=1}^mx_i}=m(m-1).$$