I posted an inequality few days ago, but I realized later on that it could only holds under stronger constraints. In particular I reduced such inequality to the inequality I am posting here. I think it is nice and I would really pleased if someone could give me any some suggestion. I simulated it with million of combinations and it always held. Thanks in advance!
Consider two tuples each of $n$ positive real numbers $(x_1, x_2, ...x_n)$ and $(y_1,y_2,...,y_n)$ and two real numbers $q \in (0,1)$ and $\beta \in (0,1)$. If $$ \sum_{i=1}^n{\frac{q}{\beta x_i}} = \sum_{i=1}^n{\frac{1-q}{(1-\beta) y_i}}, $$ is it true that the inequality below holds?$$ \sum_{i=1}^n{\frac{q}{\beta x_i}} \geq \sum_{i=1}^n{\frac{1}{\beta x_i + (1-\beta) y_i}}. $$
By Cauchy's inequality,$$ (βx_k + (1 - β)y_k)\left( \frac{q^2}{βx_k} + \frac{(1 - q)^2}{(1 - β)y_k} \right) \geqslant (q + (1 - q))^2 = 1, $$ i.e.$$ \frac{q^2}{βx_k} + \frac{(1 - q)^2}{(1 - β)y_k} \geqslant \frac{1}{βx_k + (1 - β)y_k}. \quad 1 \leqslant k \leqslant n $$
Note that$$ \sum_{k = 1}^n \frac{q}{βx_k} = \sum_{k = 1}^n \frac{1 - q}{(1 - β)y_k}, $$ thus \begin{align*} \sum_{k = 1}^n \frac{q}{βx_k} &= q \cdot \sum_{k = 1}^n \frac{q}{βx_k} + (1 - q) \cdot \sum_{k = 1}^n \frac{1 - q}{(1 - β)y_k}\\ &= \sum_{k = 1}^n \left( \frac{q^2}{βx_k} + \frac{(1 - q)^2}{(1 - β)y_k} \right)\\ &\geqslant \sum_{k = 1}^n \frac{1}{βx_k + (1 - β)y_k}. \end{align*}