# Nice inequality involving two tuples of real numbers

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?$$\sum_{i=1}^n{\frac{q}{\beta x_i}} \geq \sum_{i=1}^n{\frac{1}{\beta x_i + (1-\beta) y_i}}.$$

• Is there a typo? $q=0$ is a counterexample to both the equality and the inequality. – Stella Biderman Feb 18 '18 at 16:00
• The equality seems to be a given condition. – Saad Feb 18 '18 at 16:02
• @enricopiovano Your proposed inequality has nothing to do directly with the three inequalities you tagged, please don't misuse tags. – Saad Feb 18 '18 at 16:04
• @AlexFrancisco : thanks a lot to let me know, I am very sorry – Enrico Piovano Feb 18 '18 at 16:09
• @StellaBiderman for $q=0$, may consider $\beta=0$ and find an undertermined form. However, you are completely right, it is better to consider both $q \in (0,1)$ and $\beta \in (0,1)$. Let me modify the exercise. AlexFrancisco is completely right. the equality gives you a condition, and based on the condition you need to prove the inequality. – Enrico Piovano Feb 18 '18 at 16:15

## 1 Answer

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*}

• This is a very elegant solution ! Really thanks a lot @AlexFrancisco, very very appreciated. – Enrico Piovano Feb 19 '18 at 22:28