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Given that $x_1,x_2,x_3,y_1,y_2,y_3$ are positive real numbers satisfying

$$\frac{y_1-x_1}{x_2+x_3} + \frac{y_2-x_2}{x_3+x_1} + \frac{y_3-x_3}{x_1+x_2} > 0.$$

Show that

$$\frac{y_1-x_1}{y_2+y_3} + \frac{y_2-x_2}{y_3+y_1} + \frac{y_3-x_3}{y_1+y_2} > 0.$$

I don't think that common inequalities like AM–GM or Cauchy–Schwarz can apply since it is possible that some of the $y_i-x_i$ is negative. I don't know how to apply rearrangement inequality here. It seems that I need to divide it into many cases.

I tried to disprove this, but I can't. Otherwise, assume that $$\frac{y_1-x_1}{x_2+x_3}\le 0$$ and $$\frac{y_1-x_1}{y_2+y_3} + \frac{y_2-x_2}{y_3+y_1} + \frac{y_3-x_3}{y_1+y_2} \le 0.$$ $$\frac{y_2-x_2}{y_3+y_1} + \frac{y_3-x_3}{y_1+y_2} \le \frac{x_1-y_1}{y_2+y_3}$$ $$(y_2+y_3) \frac{y_2-x_2}{y_3+y_1} + \frac{y_3-x_3}{y_1+y_2} + y_1 \le x_1$$

If I choose $x_1$ according to this inequality and substitute some numbers into the calculator, the original condition is violated. However, I can't algebraically deduce a contradiction. Any help is appreciated.

GNUSupporter 8964民主女神 地下教會
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  • It seems like this has been asked before... – Element118 Dec 30 '15 at 10:45
  • @Element118 I forgot the keywords of the question, and I tried searching it using the MathJax code, but I can't find any results. In that post, the user didn't show his attempt. He just posts the question, and it was on hold to be closed. I don't know if it has been closed now. Anyways, I can't search that question. – GNUSupporter 8964民主女神 地下教會 Dec 30 '15 at 10:46
  • I wonder if the following is useful? What about taking the points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ as points on a triangle? In this case you would have the mean $y$ position between two of the vertices, say $1$ and $2$, as $\frac{(y_1 + y_2)}{2}$. – jim Dec 30 '15 at 11:36
  • Can you please elaborate more? I don't know how to find a meaning in coordinate geometry for the fractions in the question. – GNUSupporter 8964民主女神 地下教會 Dec 30 '15 at 11:39

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