Why is the solution of $\min_{x\ne0}\frac{\langle Ax,x\rangle}{\langle x,x\rangle}$ an eigenvector of $A$ wrt to the smallest eigenvalue? Let $A\in\Bbb R^{n\times n}$ be nonnegative and symmetric. We know that there is an orthonormal basis of $(e_1,\ldots,e_n)$ with $$A=\sum_{i=1}^n\lambda_ie_i\otimes e_i,$$ where $\lambda_1\ge\cdots\ge\lambda_k>\lambda_{k+1}=\cdots=\lambda_n=0$. Now, $$\langle Ax,x\rangle=\sum_{i=1}^n\lambda_i|\langle x,e_i\rangle|^2\tag1\;\;\;\text{for all }x\in\Bbb R^n$$ and from $(1)$ it's easy to see that a minimizer of $(1)$ over $\{x\in\Bbb R^n:\left\|x\right\|=1\}$ is given by $e_n$ and the corresponding minimum value is $\lambda_n$.

Now I've read that $x_0\in\mathbb R^n$ is an eigenvector of $A$ corresponding to the smallest eigenvalue ($\ne0$) of $A$ if and only if $x_0$ is a minimizer of $$\mathbb R^n\setminus\{0\}\ni x\mapsto\frac{\langle Ax,x\rangle}{\left\|x\right\|^2}\tag2.$$ How can we show this?

 A: We can write all vectors $x \in \mathbb{R}^n\setminus\{0\}$ in terms of $r = \|x\|$ and $y = x/\|x\|$. Notice also that this mapping from $x$ to $r,y$ is a bijection between $\mathbb{R}^n\setminus\{0\}$ and $\{r\in \mathbb{R}\ \vert\ r > 0\}\times\{y\in\mathbb{R}^n\ \vert\ \|y\|=1\}.$
Then your problem (2) can be written equivalently as
$$\min_{r>0,\ \|y\|=1} \frac{\langle Ary, ry\rangle}{\|ry\|^2}$$
and elementary simplification immediately reduces this problem to your (1). So we know that $e_n$ is a solution of (2) by your own argument.
Now for the other direction, suppose $x_0$ is a solution of (2). We can again use the above equivalence to see that $y_0 = x_0/\|x_0\|$ must be a solution of (1), and moreover we can expand $y_0$ in the eigenbasis,
$$y_0 = \sum_i \alpha_i e_i,$$
with $\sum \alpha_i^2 = 1$ since $\|y_0\|=1$.
Now 
$$\langle Ay_0, y_0\rangle = \left\langle \sum \alpha_i \lambda_i e_i, \sum \alpha_i e_i\right\rangle = \sum \lambda_i \alpha_i^2$$
by orthonormality of the eigenvector basis you've chosen for $A$.
Suppose, for contradiction, that $\alpha_i \neq 0$ for some $i \leq k$. Then 
$$\langle Ay_0, y_0\rangle = \lambda_i \alpha_i^2 + \sum_{j\neq i} \lambda_j \alpha_j^2 \geq \lambda_i \alpha_i^2 + \lambda_n (1-\alpha_i^2) > \lambda_n = \langle Ae_n, e_n\rangle,$$
contradicting the optimality of $y_0$. Therefore $y_0$ is a linear combination of the $e_{k+1}, \ldots, e_n$ and therefore $y_0$, and hence $x_0$, are eigenvectors of $A$ with eigenvalue $\lambda_n$.
A: Even if we don't know the theorem about real symmetric matrices, we can prove directly that a point $x$ where the minimal value of $\frac{\langle A x, x\rangle}{\langle x, x\rangle}$ is achieved is an eigenvector. 
The simplest way to see that: consider $x$ on the unit sphere at which $\langle A x, x\rangle $ is smallest. This is an extremum problem with one constraint, so using Lagrange multipliers: there exists $\lambda$ so that the $x$ is a critical point of function $\langle A x, x\rangle  - \lambda \langle x, x \rangle$. This means: $A x - \lambda x = 0$.  
Otherwise, we can directly calculate the derivative of $\frac{\langle A x, x\rangle}{\langle x, x\rangle}$.  At a given $x$ it is the linear function
$$y \mapsto \frac{ 2 \mathcal {Re} \langle \langle x, x\rangle A x - \langle A x, x\rangle x, y \rangle }{\|x\|^4}$$ If $x$ is a critical point, it follows that 
$$\langle x, x\rangle A x - \langle A x, x\rangle x= 0$$
This last approach is useful in the case of compact self adjoint operators on infinite dimensional Hilbert spaces.
