What will be the $(b^2-a)$?

Question

If $a$, $b$ are real numbers such that $$4a^{2}+b^2=4a-\left(\frac{1}{4b^2}\right)$$ What will be the $b^2-a$ equal to?

I tried to make this equation more basic, but I could not reach the result.

Answer 1

Rearranging we have $(2a-1)^2+(b+\frac{1}{2b})^2=2$. But $|b+\frac{1}{2b}|\ge\sqrt2$ (by AM/GM), so we must have $a=\frac{1}{2},|b|=\frac{1}{\sqrt2}$ and hence $b^2-a=0$.