Let $f(\textbf{x})=\sum_{j=1}^n\frac{a_j}{x_j}$, determine its extrema on the set $$S=\{x\in \mathbb{R}^n|\prod_{j=1}^n x_j=C\}$$
The answer is here http://math.nyu.edu/student_resources/wwiki/index.php/Advanced_Calculus:_1998_January:_Problem_1. However, I don't get the step from" Dividing $i\neq k$, we get $-\frac{a_i}{x_i}\frac{1}{\prod^n_{j\neq i}x_j}=-\frac{a_k}{x_k}\frac{1}{\prod^n_{j\neq k}x_k}$. Can someone explain this a little bit. Thanks. Or is there an easier way to go from there?
We have the system of $n$ equations written in vector form $$\left(-\frac{a_1}{x_1^2},\cdots,-\frac{a_n}{x_n^2}\right)=\lambda\left(\prod_{j\neq1}x_j,\cdots,\prod_{j\neq n}x_j\right)$$ In the step you are talking about, they have divided by the product on the RHS of the equations to give $$-\frac{a_i}{x_i^2}\frac{1}{\prod_{j\neq i}x_j}=\lambda$$ for $i=1,\cdots,n$. The RHS is common to all the equations, so we can equate all the LHS as well. So for $i\neq k$ we arrive at $$-\frac{a_i}{x_i^2}\frac{1}{\prod_{j\neq i}x_j}=-\frac{a_k}{x_k^2}\frac{1}{\prod_{j\neq k}x_j}$$ We can write this as $$\frac{a_i}{x_i}\frac{1}{\prod_{j=1}^n x_j}=\frac{a_k}{x_k}\frac{1}{\prod_{j=1}^n x_j}$$ which further gives $$\frac{a_i}{x_i}=\frac{a_k}{x_k}$$ Has this answered your question?