Let $$f(x_1, ..., x_n) = x_1 + \frac{x_2}{x_1} + \frac{x_3}{x_2} + ...+\frac{x_n}{x_{n-1}} + \frac{1}{x_n}$$ where $x_1, ..., x_n>0$.
I thought that maybe I should cast what above as optimization problem with equality constraint, but I don't know how to do that. Moreover the constraint gives unbounded open set, so even existence of extrema is tricky.
On
Clearly $f(x_1, \cdots, x_n) > x_1$ so there is no upper bound.
Assume $x_0 = x_{n+1} = 1$ and make a change of variable $y_1 = \dfrac{x_1}{x_0}$, $y_2 = \dfrac{x_2}{x_1}$, $\cdots, $ $y_{n+1} = \dfrac{x_{n+1}}{x_n}$.
Then you have the objective function $F(y_1, \cdots, y_{n+1}) = y_1 + \cdots + y_{n+1}$ under the constraint $y_1y_2\cdots y_{n+1} = 1$. This is a standard Lagrange multiplier problem (or AM-GM inequality problem).
On
You can try that , but i'm not sure if it's true or not.
$\nabla f(x_1,x_2,...,x_n)=0 $ $\iff \left(1-\dfrac{x_2}{x_1^2},\dfrac{1}{x_1}-\dfrac{x_3}{x^2_2},\cdots, \dfrac{1}{x_{n-2}}-\dfrac{x_n}{x_{n-1}^2}, \dfrac{1}{x_{n-1}}-\dfrac{1}{x_{n}^2}\right) $ $\iff x_1^2=x2,x_2^2=x_1x_3, x_3^2=x_2x_4, \cdots, x_{n-1}^2=x_{n-2}\,x_n$, $ x_n^2=x_{n-1} $
$\iff x=(x_1,x_1^2,\cdots, x_1^n)$
Since we don’t have boundary constraints, we don’t have to introduce additional Lagranges multipliers. Since $f$ is defined on an unbounded open set, necessary conditions that $f$ has an extremum at a point $x=(x_1,\dots, x_n)$ is the equality of partial derivatives of $f$ at $x$ to zero, see, for instance, [Fich, 196]. It means
$\frac{\partial f}{\partial x_1}=1-\frac {x_2}{x_1^2}=0$, so $x_2=x_1^2$.
$\frac{\partial f}{\partial x_2}=\frac 1{x_1}-\frac {x_3}{x_2^2}=0,$ so $x_3=\frac{x_2^2}{x_1}=x_1^3$.
$\dots$
$\frac{\partial f}{\partial x_i}=\frac 1{x_{i-1}}-\frac {x_{i+1}}{x_i^2}=0,$ so $x_{i+1}=\frac{ x_i^2}{x_{i-1}}=\frac{ x_1^{2i}}{x_1^{i-1}}=x_1^{i+1}$.
$\dots$
$\frac{\partial f}{\partial x_n}=\frac 1{x_{n-1}}-\frac {1}{x_n^2}=0,$ so $1=\frac{x_n^2}{x_{n-1}}=\frac{x_1^{2n}}{x_1^{n-1}}=x_1^{n+1}$.
Thus $x_1=1$, and so $x_i=1$ for each $i$.
We can further study whether $f$ has at $x$ a local minimum, a local maximum, or a saddle point by the second-derivative test, but there is an easy way.
Namely, by an inequality between an arithmetic and geometric means, for each $t=(t_1,\dots,t_n)$ we have
$$f(t)\ge (n+1)\sqrt[n+1]{t_1\cdot \frac{t_2}{t_1}\cdots\frac{t_n}{t_{n-1}}\cdot\frac1{t_n}}=n+1,$$
and the equality is attained iff
$$t_1=\frac{t_2}{t_1}=\dots=\frac{t_n}{t_{n-1}}=\frac1{t_n},$$
that is iff $$\log t_1-\log 1=\log t_2-\log t_1=\dots=\log t_n-\log t_{n-1}=\log 1-\log t_n,$$
That is when $0=\log 1,\log t_1,\dots, \log_n,\log 1=0$ are consecutive members of an arithmetic progression, that is when all $\log t_i$ are equal to zero, that is $t=x$.
Thus, the function $f$ attains a unique extremum at the point $x=(1,\dots,1)$, which is a global minimum.
References
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, vol. I, 5-th edition, M.: Nauka, 1962 (in Russian).