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Let $a_1,\dots,a_n$; $b_1,\dots,b_n$; $x_1,\dots,x_n$ and $y_1,\dots,y_n$ be positive numbers. Assume that $$\frac{\sum_{i} a_i x_i}{\sum_i b_i y_i} \leq \frac{\sum_{i} a_i^2 x_i}{\sum_i b_i^2 y_i}=1$$ I would like to show that $$\frac{\sum_{i} a_i^3 x_i}{\sum_i b_i^3 y_i}\geq 1.$$

I tried BCS inequality but could not solve it. Any ideas?

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The assertion is false. A counterexample with $n=2$ is $$ a_1=a_2=x_1=x_2=1,\quad b_1=\tfrac12,\, y_1=4,\, b_2=2,\, y_2=\tfrac14. $$ Suppose we make the change of variables $w_i = a_i^2x_i$ and $z_i = b_i^2y_i$; then the assertion becomes $$ \frac{\sum a_i^{-1}w_i}{\sum b_i^{-1}z_i} \le \frac{\sum w_i}{\sum z_i} = 1 \implies \frac{\sum a_iw_i}{\sum b_iz_i} \ge 1. $$ In this formulation, it's natural to consider the case $w_i=z_i=1$; it's easy now to see that the assertion is likely false if we take the set of $a_i$ to be closed under taking reciprocals and the set of $b_i$ to be closed under taking reciprocals.

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