Is the set
$\left\{(x_1,x_2,\dots,x_n)\in\mathbb{R}_{\geq 1}^n:\sum\limits_{i=1}^n\frac{1}{x_i}\leq 1\right\}$
convex? For $n=2$ it is not too difficult to show convexity and I guess one has to find the right transformations to avoid messy computations. I am thankful for suggestions how to solve this or where in the literature I could find the solution.
While reading about convexity today, i also came across the inequality Max Alekseyev mentioned in his answer (I wasn`t aware of it before). I found some other results and I guess it can also be proven somewhat similar to Pietro Majers comment. It can be shown that the function
$f(x)=\left(\sum\limits_{i=1}^n\frac{1}{x_i}\right)^{-1}$
is concave ($x\in\mathbb{R}_{>0}^n$). So the hypograph
$H_f=\{(x,y)\in\mathbb{R}_{>0}^n\times\mathbb{R}:y\leq f(x)\}$
is a convex set. Then
$H_f\cap\{(x,1):x\in\mathbb{R}_{>0}^n\}$
is convex and so is the projection of this set onto the first $n$ coordinates. This last set was the set from the question.
Let $$X = \left\{(x_1,x_2,\dots,x_n)\in\mathbb{R}_{\geq 1}^n:\sum\limits_{i=1}^n\frac{1}{x_i}\leq 1\right\}.$$
To prove that $X$ is convex, it is enough to show that $$\alpha\cdot x+(1-\alpha)\cdot y\in X$$ for any $x,y\in X$ and any real $\alpha\in[0,1]$.
Denote $\beta=1-\alpha$. We have $\beta\in[0,1]$ as well. It is clear that $$\alpha\cdot x+\beta\cdot y = (\alpha\cdot x_1+\beta\cdot y_1, \dots, \alpha\cdot x_n+\beta\cdot y_n)$$ and we need to show that the sum of reciprocals of these components is at most 1. For any $i=1,2,\dots,n$, we have $$\frac{1}{\alpha\cdot x_i+\beta\cdot y_i} \le \alpha\frac{1}{x_i}+\beta\frac{1}{y_i}$$ as the inequality between weighted harmonic and arithmetic means of $\frac{1}{x_i}$ and $\frac{1}{y_i}$ with the weights $\alpha$ and $\beta$, respectively. Summing up over $i$, we get $$\sum_{i=1}^n \frac{1}{\alpha\cdot x_i+\beta\cdot y_i} \le \alpha\sum_{i=1}^n\frac{1}{x_i}+\beta\sum_{i=1}^n\frac{1}{y_i} \le \alpha + \beta =1$$ as required.