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We have a box containing red and black balls. If we draw two at random the probability of getting both of them red is $1/2$. Which basically means:

\begin{equation} \frac{r}{r+b} \cdot \frac{r-1}{r+b-1} = \frac{1}{2} \end{equation} Then we have for a positive number of red and black balls, $r$ and $b$ respectively:

\begin{equation} \frac{r}{r+b} > \frac{r-1}{r+b-1} \end{equation}

and the following inequality follows:

\begin{equation} \left(\frac{r}{r+b}\right)^2 > \frac{1}{2}>\left(\frac{r-1}{r+b-1}\right)^2 \end{equation}

How can I derive this inequality?

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up vote 4 down vote accepted

If $x$ and $y$ are distinct positive numbers with $xy = z$, then one of $x$ and $y$ must be bigger than $\sqrt z$ and one must be smaller. If both $x$ and $y$ were bigger than $\sqrt z$, their product would be bigger than $z$, and if both were smaller, their product would be smaller.

So if you have $x>y$, then you must have $x > \sqrt z > y$, and so $x^2 > z > y^2$.

Here you have $x={r\over r+b}$, $y = {r-1 \over r+b-1}$ and $z=\frac12$.

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Thanks for the answer, I liked all of them, but I chose yours since I think it gave the best intuition into why it holds. – Euclean Jul 19 '12 at 14:36
You are welcome. I hoped you might appreciate an intuitive answer rather than a more symbolic one. – MJD Jul 19 '12 at 14:39

If you multiply your first inequality by $\frac {r-1}{r+b-1}$ on both sides you get the left half of your second. And if you multiply it by (what?) you get the right half of your third. If you need to derive the first, cross multiply it and it is obvious.

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So we have $x > y$ such that $x \cdot y = \frac{1}{2}$. Thus $$ x^2 = x \cdot \frac{1}{2y} \stackrel{x>y}{>} \frac{1}{2} = y \cdot \frac{1}{2 y} \stackrel{y<x \implies \frac{1}{y} > \frac{1}{x}}{>} y \cdot \frac{1}{2 x} = y^2 $$

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Take the equation

r(r-1)/{(r+b) (r+b-1)}=1/2 for comparison and then for the inequality r/(r+b) > (r-1)/(r+b-1) and multiply both sides by r/(r+b) and you get [r/(r+b)]$^2$ >r(r-1)/{(r+b) (r+b-1)}=1/2.

Now for the other side of the inequality multiple both sides by (r-1)/(r+b-1) and you have

r(r-1)/{(r+b) (r+b-1)}=1/2 > [(r-1)/(r+b-1)]$^2$.

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