Hi here is a repost of the original question, which had been closed due to "This question needs details or clarity" and then automatically deleted.
Question: Consider the function $F(x, y) = \frac{1}{y\sum_{i=0}^{n}x_i}$ with variables $x_i >0 \ \forall i$ and $y >0$. Is it a convex function? I know $\frac{1}{x_iy}$ and $\frac{1}{x_i+x_j}$ are convex functions for any $x_i, x_j >0$ and $y >0$, but how about the case where the denominator is the sum of product?
Comments of the original post:
Ted Shifrin: If $f>0$, is it true that $1/f$ is convex if and only if $f$ is concave?
@Ted Shifrin: Thanks for the comment. I think it should be "if $f>0$, $1/f$ is convex if $f$ is concave", instead of "if and only if". The function $f(x,y) = xy$ is neither convex or concave, hence the reciprocal rule does not work for this problem.
jDAQ: Did you check if $g(z,y)=\frac{1}{zy}$ is convex? Also, notice that composition with an affine mapping preserves convexity, that is, for $g$ convex you have that $f(z)=g(Az+b)$ is convex (if $Az+b$ in dom $f$ of course.)
@jDAQ: Thanks for the comment. Yes, I have checked that the hessian matrix of $g(z,y)=\frac{1}{zy}$ is positive semidefinite. But I do not understand the affine mapping here. What is $z$ and what is the specific function of $Az+b$? Could you please clarify more? Thanks.