# Inequality Equation Proof in Papoulis

I am reading Papoulis' "Probability, Random Variables, and Stochastic Processes". On page 31, equation (2-40) is derived. The implication is:

$$\left(\frac{a-1}{a+b-1}\right)^2 < \frac13 < \left(\frac{a}{a+b}\right)^2 \implies (\sqrt{3} + 1)\frac{b}{2} < a < 1 + (\sqrt{3} + 1)\frac{b}{2}$$

How was the right hand side (equation 2-40) reached?

My work: I attempted to show the right hand side using two techniques. First, by expanding the squares and doing algebraic manipulations. However, I could not remove the $a^2$ terms.

The second attempt involved taking the square root of both sides. Then, I took the reciprocal of all three sides, and reversed the direction of the inequalities. However, after I multiply by $a(a-1)$, that creates $a^2$ terms.

Any help would be appreciated. Thanks.

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While I may not be able to solve this, could you provide constraints on a and b? – NoChance Oct 16 '11 at 5:58
@EmmadKareem: Hi, $a$ and $b$ represent the number of white and black balls in a box. So, $a>0$ and $b>0$. – jrand Oct 16 '11 at 13:37
Thanks for clarification. – NoChance Oct 16 '11 at 14:19

Hint: Invert, take square roots, find upper and lower bounds for $\sqrt{3}-1$, invert again, and clear surds from the denominator.
@Dilip Sarwate: Hi, I tried the technique you mentioned, and it worked. Thanks. Eventually, the fact $\frac{1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{2}$ must be used. – jrand Oct 16 '11 at 14:10
@jrand "Eventually, the fact $\frac{1}{\sqrt{3}−1} = \frac{\sqrt{3} + 1}{2}$ must be used." That is what is I meant by "clearing surds from the denominator". It is a well-known trick: $$\frac{1}{\sqrt{3}−1} = \frac{1}{\sqrt{3}−1}\times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{\sqrt{3} + 1}{3 - 1} = \frac{\sqrt{3} + 1}{2}.$$ – Dilip Sarwate Oct 16 '11 at 15:48