Absolute conditional minimum of function in n-dimensional space

Function

$$F(x_1,x_2,...,x_n) = \sum_{i=1}^n x_i$$

on the constraint

$$G(x_1,x_2,...,x_n)=\prod_{i=1}^n x_i-1$$

what I come up with is writing down the Lagrangian:

$$L = F(x_1,x_2,...,x_n) - \lambda G(x_1,x_2,...,x_n) \implies L = \sum_{i=1}^n x_i - \lambda (\prod_{i=1}^n x_i-1)$$

taking the partial derivatives

$$\frac{dL}{dx_1} = 1 - \lambda \frac{\prod_{i=1}^n x_i}{x_1}=0 \implies \lambda \frac{\prod_{i=1}^n x_i}{x_1}=1 \implies \lambda \frac{1}{x_1}=1 \implies \lambda = x_1$$

$$\frac{dL}{dx_2} = 1 - \lambda \frac{\prod_{i=1}^n x_i}{x_2}=0 \implies \lambda \frac{\prod_{i=1}^n x_i}{x_2}=1 \implies \lambda \frac{1}{x_2}=1\implies \lambda = x_2$$

...

$$\frac{dL}{dx_n} = 1 - \lambda \frac{\prod_{i=1}^n x_i}{x_n}=0 \implies \lambda \frac{\prod_{i=1}^n x_i}{x_n}=1 \implies \lambda \frac{1}{x_n}=1 \implies \lambda = x_n$$

$$\frac{dL}{d\lambda} = - \prod_{i=1}^n x_i + 1= 0 \implies \prod_{i=1}^n x_i = 1$$

$$\lambda = x_1 = x_2 = ...=x_n = 1$$ is our critical point.

Taking the differential of our constraint

$$dG(x_1,x_2,...,x_n) = 0$$

$$\frac{\partial G}{\partial x_1}\Delta x_1+\frac{\partial G}{\partial x_2}\Delta x_2+...+\frac{\partial G}{\partial x_n}\Delta x_n=0$$

$$\frac{\prod_{i=1}^n x_i}{x_1}\Delta x_1 + \frac{\prod_{i=1}^n x_i}{x_2}\Delta x_2+...+\frac{\prod_{i=1}^n x_i}{x_n}\Delta x_n=0$$

substituting the roots of critical point $$x_1=x_2=...=x_n=1$$ and $$\prod_{i=1}^n x_i=1$$ leads to

$$\Delta x_1 + \Delta x_2 +...+ \Delta x_n = 0$$

The second order differential of Lagrangian

$$dL = \sum_{j=1}^n\sum_{i=1}^n L_{x_j x_i} \Delta x_j x_i$$

derivations showed that $$dL > 0$$ therefore the point is local minima (we are at the bottom)

currently stuck at the moment that the critical point might just be local, but have to show that it is global minimum.

$$(x_1x_2....x_n)^{1/n} \le \frac{F(x_1,...,x_n)}{n}$$.
If $$x_1x_2....x_n=1$$, then we get
$$n=F(1,...,1) \le F(x_1,...,x_n)$$.