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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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Try breaking this into the cases of $t > 0$, $t < 0$ and $t=0$. –  ncmathsadist Nov 17 '12 at 18:09
    
You should be able to follow the same process with the new version. –  Ross Millikan Nov 17 '12 at 18:30
    
@RossMillikan, thank you. –  benald Nov 17 '12 at 18:53

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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 $t \lt 0$ –  Ross Millikan Nov 18 '12 at 14:57

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