For positive $a$ and $b$, and for $u\in[0,1]$, can we solve for $v$ in $a>b\sqrt{1-v^2}+\sqrt{1-(u+v)^2}$?

Asked Jun 03 '19 at 02:57

Active Jun 03 '19 at 03:00

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Let $a$ and $b$ be positive numbers and $u \in [0, 1]$. Can we solve this inequality for $v$? $$ a > b \sqrt{1-v^2}+ \sqrt{1-(u+v)^2} $$ I need the points for $v$ in the interval $[-1, 1-u]$ for which the corresponding equality holds, together with the information on which side of a point the inequality holds.

When I enter the corresponding equality in Mathematica, I get a ridiculously large formula. Any ideas on how to do this manually?

Context: I'm trying to expand this answer of mine. And I think solving this can help me.

edited Jun 03 '19 at 03:00

Blue

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asked Jun 03 '19 at 02:57

Paul

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A friend just told be how to do this, squaring twice, the second time move the square root to one side. – Paul Jun 03 '19 at 17:37
Then you have a fourth degree polynomial – Paul Jun 03 '19 at 23:36

For positive $a$ and $b$, and for $u\in[0,1]$, can we solve for $v$ in $a>b\sqrt{1-v^2}+\sqrt{1-(u+v)^2}$?

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