Inscribed Circle !

Let $P: (x,\dfrac{1}{1 + x^2})$ and $O: (0,0)$

To minimize $\overline{OP} :$

$D = (\overline{OP})^2 = x^2 + (1 + x^2)^{-2} \implies$ $\dfrac{dD}{dx} = 2x(\dfrac{(1 + x^2)^3 - 2}{(1 + x^2)^3}) = 0$

$x \neq 0 \implies x = \pm\sqrt{2^{\frac{1}{3}} - 1} \implies$ $y = \dfrac{1}{2^{\frac{1}{3}}} \implies$ $OP = \dfrac{\sqrt{3 - 2^{\frac{2}{3}}}}{2^{\frac{1}{3}}}$

$\implies$ The area of the inscribed circle $A = \pi (OP)^2 =$ $\dfrac{3 - 2^{\frac{2}{3}}}{2^{\frac{2}{3}}}\pi$

$= \dfrac{a - b^{\frac{b}{a}}}{b^{\frac{b}{a}}}\pi \implies a + b = \boxed{5}$ .

Note:

$0 < x < \sqrt{2^{\frac{1}{3}} - 1} \implies \dfrac{dD}{dx} < 0$ and $x > \sqrt{2^{\frac{1}{3}} - 1} \implies \dfrac{dD}{dx} > 0$

$\implies$ relative min at $x = \sqrt{2^{\frac{1}{3}} - 1}$