A Controversial Question :P

Why It is controversial Question ?

Let $m$ be the slope at any point of the circunference:

$m=\dfrac{dy}{dx}=-\dfrac{d/dx}{d/dy}=-\dfrac{2x}{2y}=-\dfrac{x}{y}$

Now, let the point of tangency be $P=(a,b)$ , and let $OB=x,0)$ and $OA=(0,y)$ . Then we have $a^2+b^2=4$ , and let's find the equation of the line passing through $P$ with slope $m$ :

$y-b=m(x-a) \implies y-b=-\dfrac{a}{b}(x-a) \implies ax+by=a^2+b^2 \\ ax+by=4$

Now, find $OB$ :

$y=0 \implies ax=4 \implies x=\dfrac{4}{a} \implies OB=\left(\dfrac{4}{a},0\right)$

And find $OA$ :

$x=0 \implies by=4 \implies y=\dfrac{4}{b} \implies OA=\left(0,\dfrac{4}{b}\right)$

So, $OA+OB=\dfrac{4}{a}+\dfrac{4}{b}$ . Finally, using the QM-HM inequality we have:

$\sqrt{\dfrac{a^2+b^2}{2}} \geq \dfrac{2}{\frac{1}{a}+\frac{1}{b}} \\ \sqrt{2} \geq \dfrac{2}{\frac{1}{a}+\frac{1}{b}} \\ \dfrac{1}{a}+\dfrac{1}{b} \geq \sqrt{2} \\ \dfrac{4}{a}+\dfrac{4}{b}\geq 4\sqrt{2} \\ OA+OB \geq \boxed{4\sqrt{2}}$