Exe 3.8 Sorry, it is a problem that appears in Jehle and Reny Advanced Microeconomic Theory (3rd ed) exercise 3.8. But I think it's a partial derivative question.
Letting $f_i(\mathbf{x})=\partial f(\mathbf{x})/\partial x_i$, ($\mathbf{x}$ is a vector, a commodity bundle, and $x_i$ is a scalar, commodity $i$ in the bundle).
Show that (this is my main question.),
$\sigma_{ij}(\mathbf{x})\equiv -\frac{x_if_i(\mathbf{x})+x_jf_j(\mathbf{x})}{f^2_j(\mathbf{x})f_{ii}(\mathbf{x})+2f_i(\mathbf{x})f_j(\mathbf{x})f_{ij}(\mathbf{x})+f_i^2(\mathbf{x})f_{jj}(\mathbf{x})}\frac{f_i(\mathbf{x})f_j(\mathbf{x})}{x_ix_j}$
Then using the above formula, show that $\sigma_{ij}(\mathbf{x})\geq0$ whenever $f$ is increasing and concave.
PS: I know $\sigma_{ij}$ can be written as
$\sigma_{ij}=-\frac{d (x_i/x_j)}{x_i/x_j} \frac{f_i(\mathbf{x})/f_j(\mathbf{x})}{d (f_i(\mathbf{x})/f_j(\mathbf{x}))}$
How can we derive the result from this known fact? Thank you.
Is it assumed that $x_i \geq 0$ for each $i$? I think we'd have to or else we'd be in trouble. For now I'm assuming $x_i \geq 0$, so that makes $x_ix_j$ positive.
Since $f$ is increasing $f_i(\boldsymbol{x}) \geq 0$ for each $i$. That makes $x_if_i(\boldsymbol{x})+x_jf_j(\boldsymbol{x})$ positive. Finally, $f$ being concave means NEGATIVE second order (partial) derivatives, so \begin{align*} f^2_j(\boldsymbol{x})f_{ii}(\boldsymbol{x})+2f_i(\boldsymbol{x})f_j(\boldsymbol{x})f_{ij}(\boldsymbol{x})+f^2_i(\boldsymbol{x})f_{jj}(\boldsymbol{x}), \end{align*} is NEGATIVE. That gives us \begin{align*} -\frac{+}{-}\frac{+}{+}=+. \end{align*} Hope that helps.