Proof of Concavity of Cobb-Douglas Function

Question

Given the function $F(\mathbf{x})=x^{a_1}_1x^{a_2}_2 \ldots x^{a_n}_n$ defined on the set $S=\{\mathbf{x}=(x_1, \ldots, x_n) \in \mathbb{R}^n: x_1>0, \ldots ,x_n>0\}$ with $a_1,a_2,\ldots,a_n > 0$ and $a_1+a_2+\ldots+a_n=3$, I want to:

(i) Show that $\mathbf{x} \cdot \nabla F(\mathbf{x}) = 3F(\mathbf{x})$ at every $\mathbf{x}$, where $\nabla F(\mathbf{x}) = (\frac{\delta F(\mathbf{x})}{\delta x_1},\ldots,\frac{\delta F(\mathbf{x})}{\delta x_n})$.

I was able to work out that $\mathbf{x} \cdot \nabla F(\mathbf{x}) = a_1x^{a_1}_1 + \ldots + a_nx^{a_n}_n$ but got no further. Need help with this part!

(ii) Determine whether $F(\mathbf{x})$ is concave in $\mathbf{x}$ on the set $\mathbf{x}$.

My first thought was to use a Hessian matrix but that would be too tedious for this function. Is there a better method?

Answer 1

If you look at the case $x_1=x_2=\ldots x_n=t>0$, you have $$F(x_1,x_2,\ldots,x_n)=t^{a_1}\cdot t^{a_2}\cdots t^{a_n}.$$ Verify that the function $t\mapsto t^{a_1}\cdot t^{a_2}\cdots t^{a_n}$ is not concave. This implies that $F$ is not concave either.