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Suppose $$ a^k(1-a)^{n-k}=b^m(1-b)^{n-m},$$ where $0<a,b<\frac{1}{2}$ are real numbers, $n,k,m$ are positive integers, $0<k<m<n$.

How to prove that $a<b$?

I am feeling this should be trivial, but somehow I am stuck...

(This is related to this question, by the way)

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$a<b$ may not be verified. Let $k=1,m=2$ and let $n$ moves. Then, we will study : $$a(1-a)^{n-1}=b^2(1-b)^{n-2} \Leftrightarrow \frac a b \left( \dfrac {1-a}{1-b}\right)^{n-1}=\dfrac b {1-b}.$$

But, we have the following :

$$\forall a,b \in (0,1/2) \text{ s.t. } 0<a<b \implies \dfrac{1-a}{1-b}>1.$$

Then : $$\underset{n \to \infty} \lim \dfrac a b \left( \dfrac {1-a}{1-b}\right)^{n-1}=\infty.$$

Hence, if $n$ is big enough, the equality does not occur since $\frac b {1-b} <1$.

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  • $\begingroup$ Concretely, if $a = .3, b = .2, k = 1, m = 2$, then $\frac{a^k(1-a)^{n-k}}{b^m(1-b)^{n-m}} \approx 1.057...$ for $n = 14$ and $\approx 0.9252...$ for $n = 15$. So nothing can be concluded about the relation of $a$ and $b$. $\endgroup$ – Hans Engler Aug 29 '17 at 16:27

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