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I am working on research involving probability tables. I simplified the problem to the following. Say we have the following:

$C_1, C_2, C_3, C_4$

$\forall i, C_i > 0$

$x = \frac{C_1 + C_2}{C_1 + C_2 + C_3 + C_4}$

$y = \frac{C_1}{C_1 + C_3}$

$z = \frac{C_2}{C_2 + C_4}$

I am almost positive (by just filling in numbers) that both $y$ and $z$ cannot $> x$ or $< x$. Meaning that unless $y = z$, then either $y < x$ and $z > x$ or $y > x$ and $z < x$. Does anyone have suggestions for an approach to prove this?

==EDIT== Basically, I am trying to prove that either of these two situations are impossible:

$z \geq y > x$

$x > y \geq z$

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The number $x$ defined in this way is called the mediant of the two numbers $y$ and $z$ defined in this way. It is known that the mediant of two distinct numbers is strictly between those numbers. See this question and its answers.

(This answer replaces my previous answer completely, as that answer was based on a slight misformatting of the question, which has since been corrected.)

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    $\begingroup$ I'm not quite sure I follow, in that example $x=\frac{4}{7}$, $y=\frac{2}{4}$ and $z=\frac{2}{3}$. In that case, $z > x > y$, which is what I believe to hold true. I am trying to prove that it is impossible for $x > z \geq y$ or $z > y \geq x$ $\endgroup$
    – Jeremiah Chen
    Nov 16, 2014 at 3:50
  • $\begingroup$ Sorry about the unclear formatting - just fixed it! $\endgroup$
    – Jeremiah Chen
    Nov 16, 2014 at 4:03
  • $\begingroup$ OK, fixed the answer to correspond this time (I hope) to the question you meant to ask! $\endgroup$
    – David K
    Nov 16, 2014 at 4:11
  • $\begingroup$ Great thanks - would like to vote your answer up but I don't have enough reputation =T $\endgroup$
    – Jeremiah Chen
    Nov 16, 2014 at 4:15
  • $\begingroup$ That's OK, I decided I don't need any more credit for this since all I really did in the end was remember I'd seen a question about the mediant before, which enabled me to look up someone else's answer. Glad to help, anyway. $\endgroup$
    – David K
    Nov 16, 2014 at 4:23

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