Extrema of function with lagrange multipliers

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

$$f(t)ge (n+1)sqrt[n+1]{t_1cdot frac{t_2}{t_1}cdotsfrac{t_n}{t_{n-1}}cdotfrac1{t_n}}=n+1,$$

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).

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} $

$iff x=(x_1,x_1^2,cdots, x_1^n)$

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 $y_1y_2cdots y_{n+1} = 1$. This is a standard Lagrange multiplier problem (or AM-GM inequality problem).