Given that $x,y\in[0,1]$, find the minimum of the following $$f(x,y)=\sqrt{x+y}+\sqrt{(1-x)+y}+\sqrt{x+(1-y)}+\sqrt{(1-x)+(1-y)}$$
I have found $$f(0,0)=f(0,1)=f(1,0)=f(1,1)=2+\sqrt{2}, f(\dfrac{1}{2},\dfrac{1}{2})=2+\sqrt{6}$$ so is $2+\sqrt{2}$ the minimum? How to prove this?
$\frac{\partial^2f}{\partial x^2}=\frac{\partial^2f}{\partial y^2}=-\frac{1}{4}\left((x+y)^{-\frac{3}{2}}+(1-x+y)^{-\frac{3}{2}}+(1-y+x)^{-\frac{3}{2}}+(2-x-y)^{-\frac{3}{2}}\right)<0$,
which says that $f$ is a concave function of $x$ and of $y$.
Id est, $\min\limits_{\{x,y\}\subset[0,1]}f=\min\limits_{\{x,y\}\subset\{0,1\}}f=f(0,0)=2+\sqrt2$