General Inequalities

I think the answer is 1/2.

This is not the answer you need but

if y=0 and x=1 (or vise versa) then the function value is 1

So the minimum value is greater or equal to 1.

Make use of the symmetry between x and y here.  The maximum (and minimum) of the function will occur at a point where x = y, hence will occur at the maximum of the function \(f(x)=2x\sqrt{1-x^2}\)

This occurs when \(x = \frac{\sqrt2}{2}\) and \(f(\sqrt2/2)=1\)