Show that for $a,b \in \mathbb{R}$ such that $2b>1$ and $a>\sqrt{b}$ the following equality is true:


I started as follows


I do not know what to do with this. I would be grateful for any help.


1 Answer 1


That's the same as $a^4-b^2 > a^2-b$ or $(a^2-b)(a^2+b) > a^2-b$ which makes it obvious.

Just in case:

$a^2+b > 2b > 1$.


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