# A strange inequality

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?

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.