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May 1st, 2017, 04:43 AM   #1
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inequality

If for all x,y and a>0 we have ;

$\displaystyle \sqrt{x^2+y^2}<a$ then find a b>0 such that:

$\displaystyle \sqrt{(z-x)^2+(w-y)^2}<b\Longrightarrow\sqrt{z^2+w^2}<a$ for all z,w
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