Show that $f(x,y) = (x^{\alpha} + y^{\alpha})^{1/\alpha}$ is concave for $0 < \alpha < 1$ and for $\alpha < 0$

Question

I am having difficult in showing that the function $f: \mathbb{R}_+ \times \mathbb{R}_+ \rightarrow \mathbb{R}$ defined as above is concave. Have tried to show that the functions $x \mapsto x^\alpha$ and $x \mapsto x^{1/\alpha}$ have good properties and also tried to verify $f$ Hessian matrix, but this has resulted lots of calculation.

Answer 1

Actually it is simple to evaluate the Hessian by hand which is given by $$H(x, y) = - \frac{(1-\alpha)x^\alpha y^\alpha(x^\alpha + y^\alpha)^{1/\alpha}}{x^2y^2(x^\alpha + y^\alpha)^2} \left( \begin{array}{cc} y^2 & -xy \\ -xy & x^2 \\ \end{array} \right). $$ Since $\left( \begin{array}{cc} y^2 & -xy \\ -xy & x^2 \\ \end{array} \right) = [y, -x]^T[y, -x]$ is positive semidefinite, $H(x, y)$ is negative semidefinite. We are done.