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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?

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    $\begingroup$ To check whether $F$ is concave or not, you could take a look at the case $x_1=x_2=\ldots=x_n=t>0$. Then you get a function of a single argument $t$, and this will turn out to be enough for checking whether $F$ is concave on $S$ (not on $x$!) or not. $\endgroup$ – Michael Greinecker Mar 1 at 14:21
  • $\begingroup$ @MichaelGreinecker consider expanding your comment into an answer $\endgroup$ – 1muflon1 Mar 1 at 14:29
  • $\begingroup$ @1muflon1 Done. $\endgroup$ – Michael Greinecker Mar 1 at 14:42
  • $\begingroup$ Check $\mathbf{x} \cdot \nabla F(\mathbf{x})$ again, you seem to have "differentiated away" the constants in a product, which gave you the wrong result. $\endgroup$ – VARulle Mar 1 at 21:49
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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.

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    $\begingroup$ Alternatively, recognize that the way the question is formulated implies that the answer doesn't depend on $n$. Then choose $n=1$ for simplicity and the answer is obvious. (Or, if not, calculate the second derivative.) $\endgroup$ – VARulle Mar 1 at 21:54

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