Squares inscribed in a Circle

$(a + b)^2 + a^2 = 1 \implies b^2 + 2ab + 2a^2 - 1 = 0 \implies b = \sqrt{1 - a^2} - a$ , dropping the negative root

$\implies A = a^2 + b^2 = a^2 + (\sqrt{1 - a^2} - 1)^2 = a^2 - 2a\sqrt{1 - a^2} + 1 \implies$

$\implies \dfrac{dA}{da} = \dfrac{a\sqrt{1 - a^2} - (1 - 2a^2)}{\sqrt{1 - a^2}} = 0 \implies$ $a^2(1 - a^2) = (1 - 2a^2)^2$

$\implies 4a^4 - 4a^2 + 1 = a^2 - a^4 \implies 5a^4 - 5a^2 + 1 = 0 \implies a^2 = \dfrac{5 \pm \sqrt{5}}{10}$ $\implies a = \sqrt{\dfrac{5 \pm \sqrt{5}}{10}}$

For $(+) \implies b = \sqrt{\dfrac{5 - \sqrt{5}}{10}} - \sqrt{\dfrac{5 + \sqrt{5}}{10}} < 0 \therefore$

$a = \sqrt{\dfrac{5 - \sqrt{5}}{10}} \implies b = \sqrt{\dfrac{5 + \sqrt{5}}{10}} - \sqrt{\dfrac{5 - \sqrt{5}}{10}}$

$\implies b^2 =\dfrac{\sqrt{5} - 2}{\sqrt{5}} = \dfrac{5 - 2\sqrt{5}}{5}$ and $a^2 = \dfrac{5 - \sqrt{5}}{10} \implies$

The $min(a^2 + b^2) = \boxed{\dfrac{3 - \sqrt{5}}{2}}$

and $\dfrac{a}{b} = \sqrt{\dfrac{5 - \sqrt{5}}{2(5 - 2\sqrt{5})}} = \sqrt{\dfrac{(5 - \sqrt{5})(5 + 2\sqrt{5})}{2 * 5}} =$

$\sqrt{\dfrac{3 + \sqrt{5}}{2}} = \sqrt{(\dfrac{\sqrt{5} + 1}{2})^2} = \boxed{\dfrac{\sqrt{5} + 1}{2}}$

$\implies min(a^2 + b^2) + \dfrac{a}{b} = \boxed{2}$ .

Note: You can check that a min does occur at $a = \sqrt{\dfrac{5 - \sqrt{5}}{10}}$ .

$-\sqrt{\dfrac{5 - \sqrt{5}}{10}} < a < \sqrt{\dfrac{5 - \sqrt{5}}{10}} \implies \dfrac{dA}{da} < 0$ choosing $a = 0$

and

$\sqrt{\dfrac{5 - \sqrt{5}}{10}} < a < \sqrt{\dfrac{5 + \sqrt{5}}{10}} \implies \dfrac{dA}{da} > 0$ choosing $a = \dfrac{3}{4}$