From griva,nash,sofer ex 3.10 page 53.
Given the function $f(x_1,x_2)= \alpha x_1^p x_2^q$ on $S=\{x:x>0\}$, say for which values of $\alpha,p,q$ the function is convex/strictly convex, concave/strictly concave.
I first tried computing the Hessian $\nabla^2 f$ getting the matrix
\begin{bmatrix}\alpha p(p-1)x_1^{p-2}x_2^q & \alpha pq x_1^{p-1}x_2^{q-1}\\\alpha pq x_1^{p-1}x_2^{q-1}&\alpha q(q-1)x_1^px_2^{q-2}\end{bmatrix}
Then I tried to study the sign of $\begin{bmatrix} y_1 & y_2 \end{bmatrix} \nabla^2f \begin{bmatrix} y_1 \\\ y_2\end{bmatrix}$ but here I get stuck, not knowing how to decompose the resulting polynomium.
I also tried with the basic definition: $f$ is convex if and only if \begin{matrix} f(\alpha x + (1- \alpha)y) \leq \alpha f(x) + (1- \alpha) f(y) \end{matrix} , using the corresponding definition for the (strictly) concave case.
Thus: $f(\lambda x + (1 - \lambda) y) = \alpha (\lambda x_1 + (1-\lambda)y_1)^p (\lambda x_2 + (1 - \lambda)y_2)^q $. If $p$ and $q$ are even, for the convexity of $x^p$ and $x^q$ I can write
\begin{matrix} \alpha (\lambda x_1 + (1-\lambda)y_1)^p (\lambda x_2 + (1 - \lambda)y_2)^q &\leq \alpha (\lambda x_1^p + (1-\lambda)y_1^p) (\lambda x_2^q + (1 - \lambda)y_2^q) \end{matrix}
but i don't know how to get from here to the form $\leq \lambda f(x) + (1-\lambda)f(y) )$ to prove convexity
