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I have the following question:

Given a line $y=\theta(1-x)$ where $0<x<1$, $0<y<1$ and $0<\theta<1$, I have a collection of curves $$ y^K=1-(1-x)^K $$ parametrized by an integer $K>2$. For any $K$, the related curve intersects the line at only one point. For this intersection point let $$ A=y^K $$ On the other hand I have \begin{equation} \prod_{i=1}^K {y_i}=1-\prod_{i=1}^K(1-x_i) \end{equation} where $x_i$ and $y_i$ are some points on $y=\theta(1-x)$ and let $$ B=\prod_{i=1}^K {y_i} $$ Is it possible that $B<A$?

Thank you very much in advance for any constructive comment.

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up vote 1 down vote accepted

Replace $1-x$ by $x$. Then you have the line $\ \ell\colon\ y=\theta x$ and for a given $K>2$ the curve $$\gamma: \quad x^K+y^K=1\ ,$$ where everything is restricted to the window $[0,1]^2$. The curve $\gamma$ intersects $\ell$ in a point $p:=(u,v)$. Obviously $v=\theta u$, and as $p\in\gamma$ we get the equation $$u^K(1+\theta^K)=1\ .$$ This implies $$A:=v^K=\theta^K u^K={\theta^K\over 1+\theta^K}\ .$$ On the other hand you are given $K$ points $(x_i,y_i)\in\ell$, chosen such that $$\prod_{i=1}^K x_i + \prod_{i=1}^K y_i=1\ .$$ As $y_i=\theta x_i$ for each $i$ we have $(\theta^{-K}+1)\prod_{i=1}^K y_i=1$ or $$B:=\prod_{i=1}^K y_i={1\over \theta^{-K}+1}={\theta^K\over 1+\theta^K}=A\ .$$

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Thank you very much for the solution.. – Seyhmus Güngören Jul 27 '12 at 10:28

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