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In the following inequality: $1+(y+s)^{2}>t(1+(y-s)^{2})$,$t\in{R},s>0.$How can I obtain the regions for y? Sorry, I modified the second power due to mistake. $$ Thank you!

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    @RossMillikan, thank you.2012-11-17

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Note: this is for an earlier version of the problem, but the same approach appllies.

Assuming you are treating $s,t$ as given, you can expand it to $1+y+s^2 \gt t +ty^2 - 2sty +ts^2$ then collect the $y$'s on one side to get $1+s^2 \gt t-y+ty^2-2sty+ts^2 \\1+s^2 \gt t(y^2-2sy-\frac yt)+ts^2+t$ complete the square to get $1+s^2 \gt t( y-s-\frac 1{2t})^2+ts^2+t-s^2-\frac 1{4t^2}$ and put the knowns (everything else) on the other $1+2s^2-ts^2-t+\frac 1{4t^2}\gt t( y-s-\frac 1{2t})^2$ So $y$ will be in an interval around $s+\frac 1t$

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    Note that if you divide by $t$ you need to reverse the inequality when $t \lt 0$2012-11-18