Fraction weights and weighted average smaller than sum op parts

Question

I am considering a weighted average $\bar{x} = \frac{w_1x_1+w_2x_2}{w_1+w_2}$, where $x_1<x_2$. I have seen with a lot of numerical experiments that the following relation must hold: $\frac{w_1+w_2}{\bar{x}} < \frac{w_1}{x_1} + \frac{w_2}{x_2}$. However, I am not able to formally prove it. Could someone help me out?

Answer 1

Yes, it's true, assuming $w_1$, $w_2$, $x_1$, and $x_2$ are all positive.

We know that $(x_1 - x_2)^2 > 0$, since $x_1 \neq x_2$. So $$ \begin{align*} 0 &< x_1^2 - 2x_1 x_2 + x_2^2\\ 2 x_1 x_2 &< x_1^2 + x_2^2 \\ 2 w_1 w_2 x_1 x_2 &< w_1 w_2 (x_1^2 + x_2^2)\\ (w_1^2 + 2 w_1 w_2 + w_2^2) x_1 x_2 &< w_1^2 x_1 x_2 + w_1 w_2 (x_1^2 + x_2^2) + w_2^2 x_1 x_2 \\ (w_1 + w_2)^2 x_1 x_2 &< (w_1 x_2 + w_2 x_1)(w_1 x_1 + w_2 x_2) \\ \frac{(w_1 + w_2)^2}{w_1 x_1 + w_2 x_2} &< \frac{w_1 x_2 + w_2 x_1}{x_1 x_2} \\ \frac{w_1 + w_2}{\bar{x}} &< \frac{w_1}{x_1} + \frac{w_2}{x_2} \end{align*} $$ bearing in mind that $$ \frac{w_1 + w_2}{\bar{x}} = \frac{w_1 + w_2}{\frac{w_1x_1 + w_2 x_2}{w_1 + w_2}} = \frac{(w_1 + w_2)^2}{w_1 x_1 + w_2 x_2}. $$