Min-Max Principle $\lambda_n = \inf_{X \in \Phi_n(V)} \{ \sup_{u \in X} \rho(u) \}$ - Explanations In general, I am generally someone who like to solve questions with visual support. With this idea in mind, is it someone could explain to me, with a visual support if possible, how is it possible to interpret the Min-Max principle? You could possibly find more information on the subject page $3$ (Minimax Principle, Version 2) of the document Weyl's law.

Lemma. (Minimax Principle, Version 2) $$\lambda_n = \inf_{X \in \Phi_n(V)} \sup_{u \in X} \rho(u)$$ where
  $\Phi_n(V)=\{n-\text{dimensional linear subspace } X \subset V \}$.

Precision : $\rho(u)$ is the Rayleigh quotient
 A: A simple example
Let us look at a specific example in two dimensions (the main point would probably come forward better in three dimensions, but that is harder to draw, and is left as an exercise to you in the end).
We study, as an example, the symmetric matrix
$$
A=
\begin{bmatrix}
4 & -2\\
-2 & 7
\end{bmatrix}.
$$
A standard calculation shows that it has the smallest eigenvalue $\lambda_1=3$ with normalized eigenvector $u_1=\frac{1}{\sqrt{5}}(2,1)$ and the largest eigenvalue $\lambda_2=8$ with normalized eigenvector $u_2=\frac{1}{\sqrt{5}}(-1,2)$.
Let us consider the following picture.

Memories from linear algebra
We first concentrate on the left picture. It shows a unit vector rotating (painting the black circle) with its image after multiplication by $A$ (painting the purple ellipse). Note that the semi-axes of the ellipse correspond to the eigenvectors ($\pm u_1$ and $\pm u_2$). In fact, the smaller semi-axis is $3$ and the larger is $8$. I hope this kind of picture is well-known from a course in linear algebra.
Rayleigh quotient view
So how does this correspond to the Rayleigh quotients and the min-max statement? We consider first, the statement for $n=1$. Then we are supposed to consider subspaces $X$ of dimension 1. Note that each unit vector is a basis vector of such a subspace $X$. Moreover, the Rayleigh quotient is constant in each such subspace so it suffices to study the unit vectors (this you probably know?), since, if $u$ is a unit vector and $t$ a scalar,
$$
\rho(tu)=
\frac{\langle Atu,tu\rangle}{\langle tu,tu\rangle}=
\frac{\langle Au,u\rangle}{\langle u,u\rangle}=
\langle Au,u\rangle
$$
This means that the Rayleigh quotient is constant in each such subspace $X$, and we just have to find the subspace (direction!) in which the Rayleigh quotient is minimal (the infimum part of the inf-sup).
In the figure, to the right, I have drawn the value of the Rayleigh quotient $\rho(u)=\langle Au,u\rangle$, corresponding to the unit vector $u$ used to the left. If you stare long enough on those two pictures, you will see that the Rayleigh quotient has its minimum value precisely when $u$ equals $u_1$ (and $-u_1$). I have marked these minima with red dots, and if you look at the value, you see that it is exactly $3$! This, in fact, solves our problem of finding $\lambda_1$!
We turn to $\lambda_2$. Now, we are supposed to look at two-dimensional subspaces. In our case, this means that we should look at the full space $\mathbb R^2$. The $\sup_{u\in X}\rho(u)$ becomes $\sup_{u\in\mathbb R^2}\rho(u)$, and by looking at our graph to the right, we find that the maximum is $8$, marked green. By no surprise, it is attained at $u=\pm u_2$. In this case, the inf in the inf-sup becomes trivial, since we only have one subspace, $\mathbb R^2$. We conclude that $\lambda_2=8$.
Some comments
This example is very simple, but it hopefully gives an idea. I did not mention above, but if $u=(x,y)$ then
$$
\rho(u)=\langle Au,u\rangle=4x^2-4xy+7y^2,
$$
a quadratic form! If you want to perform the calculations yourself, you can for example use the parametrization $x=\cos t$ and $u=\sin t$ and work with what you get.
In the case of higher dimensions it is of course more complicated.
Exercise
Work out the case
$$
A=
\begin{bmatrix}
2 & 0 & 0\\
0 & 7 & -1\\
0 & -1 & 7
\end{bmatrix}
$$
and think about the geometrical picture in $\mathbb R^3$.
Infinite-dimensional exercise
Try to understand what is going on in the Dirichlet eigenvalue problem
$$
\left\{
\begin{aligned}
-u''(x)&=\lambda u(x)\quad 0<x<\pi\\
u(0)&=0\\
u(\pi)&=0.
\end{aligned}
\right.
$$
You might want to use the fact that the set of eigenfunctions $\{\sin k x\}_{k=1}^{+\infty}$ is complete in $L^2((0,\pi))$.
