Cómo probar $\sum_{j=1}^{n}\frac{w_{j}}{1-p_{j}}\cdot\sum_{j=1}^{n}\frac{w_{j}}{1+p_{j}}\le \left(\sum_{j=1}^{n}\frac{w_{j}}{1-p^2_{j}}\right)^2$

Question

Deje $w_{j}\ge 0,j=1,2,3,\cdots,n,n\ge 2$ y tal $$w_{1}+w_{2}+\cdots+w_{n}=1$$ para cualquier $p_{j}\in[0,1),j=1,2,3,\cdots,n)$ tienen $$\sum_{j=1}^{n}\dfrac{w_{j}}{1-p_{j}}\cdot\sum_{j=1}^{n}\dfrac{w_{j}}{1+p_{j}}\le \left(\sum_{j=1}^{n}\dfrac{w_{j}}{1-p^2_{j}}\right)^2$$

este problema es de la escuela secundaria Shang hai in china test.maybe this problem can use integral inequality to solve it? or others.Thank you

Answer 1

Esto es simplemente AM-GM:

$$\sum \frac{w_j}{1 - p_j} \cdot \sum \frac{w_j}{1 + p_j} \le \left( \frac{\sum \frac{w_j}{1 - p_j} + \sum \frac{w_j}{1 + p_j}}{2} \right)^2 = \left(\sum \frac{w_j}{1 - p_j^2}\right)^2.$$