For instance, $f(x)=\sqrt{x}$ is clearly strictly concave in $\mathbb{R}_+$ but if we consider that function in two dimensions, i.e. $f(x,y)=\sqrt{x}$ with $(x,y)\in\mathbb{R}^2_+$, it seems that it is not strictly concave anymore. I.e, for any two points $(x,y_1)$ and $(x,y_2)$ we have that $$ f(\lambda x + (1-\lambda) x,\lambda y_1 + (1-\lambda) y_2)=\sqrt{x}=\lambda f(x,y_1)+(1-\lambda)f(x,y_2) $$ so it is not strictly concave. Is this reasoning correct?
If so, I am having trouble reconciling this fact with the definition of strict concavity through the Hessian. Considering the function $f(x,y)=\sqrt{x}$ on $\mathbb{R}^2_+$, the Hessian approach yields $$ \left[\begin{array}{cc} x & y\end{array}\right]H\left[\begin{array}{c} x\\ y \end{array}\right]=\begin{matrix}\left[\begin{array}{cc} x & y\end{array}\right]\left[\begin{array}{cc} f_{xx}(x,y) & 0\\ 0 & 0 \end{array}\right]\left[\begin{array}{c} x\\ y \end{array}\right]=x^{2}f_{xx}(x,y)\end{matrix} $$ so that $H$ is negative definite for $x>0$ which means that $f$ should be strictly concave.
