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Let's go through the problem presented in the image step by step.

### Given Information:
- Let \( a \in [0; 1] \).
- We define the sequence \( (u_n) \) by \( u_0 = a \) and for all \( n \in \mathbb{N}, \ u_{n+1} = 3u_n(1 - u_n) \).

### Tasks:

1. **Prove that for all \( x \in [0; 1] \), we have:**
   \[
   x(1 - x) \leq \frac{1}{4}.
   \]
   
2. **Prove that for all \( n \in \mathbb{N} \), we have:**
   \[
   u_n \in [0; 1].
   \]

### Solution Outline:

#### 1. Proving \( x(1 - x) \leq \frac{1}{4} \) for \( x \in [0; 1] \):

Let \( f(x) = x(1 - x) \) for \( x \in [0, 1] \).

- First, we calculate the derivative of \( f(x) \):
  \[
  f'(x) = 1 - 2x.
  \]
- Setting \( f'(x) = 0 \), we find the critical point:
  \[
  1 - 2x = 0 \implies x = \frac{1}{2}.
  \]
- Now, evaluate \( f(x) \) at \( x = 0 \), \( x = 1 \), and \( x = \frac{1}{2} \):
  \[
  f(0) = 0, \quad f(1) = 0, \quad f\left( \frac{1}{2} \right) = \frac{1}{4}.
  \]
- Since \( f(x) \) is a quadratic function opening downwards and attains its maximum value at \( x = \frac{1}{2} \), the maximum value of \( f(x) = x(1 - x) \) on the interval \( [0, 1] \) is \( \frac{1}{4} \).

Therefore, for all \( x \in [0, 1] \),
\[
x(1 - x) \leq \frac{1}{4}.
\]

#### 2. Proving \( u_n \in [0; 1] \) for all \( n \in \mathbb{N} \):

- **Base case (for \( n = 0 \)):** We are given that \( u_0 = a \in [0; 1] \), so the base case holds.

- **Inductive step:** Assume \( u_n \in [0; 1] \) for some \( n \geq 0 \). We need to prove that \( u_{n+1} \in [0; 1] \).
  
  From the recurrence relation:
  \[
  u_{n+1} = 3u_n(1 - u_n).
  \]
  
  - If \( u_n = 0 \), then \( u_{n+1} = 0 \).
  - If \( u_n = 1 \), then \( u_{n+1} = 0 \).
  
  Now, consider \( u_n \in (0, 1) \). We know from part 1 that for \( u_n \in [0, 1] \),
  \[
  u_n(1 - u_n) \leq \frac{1}{4}.
  \]
  
  Therefore, 
  \[
  u_{n+1} = 3u_n(1 - u_n) \leq 3 \times \frac{1}{4} = \frac{3}{4}.
  \]
  
  Since \( u_{n+1} \geq 0 \) and \( u_{n+1} \leq \frac{3}{4} \), we have \( u_{n+1} \in [0, 1] \).

By induction, \( u_n \in [0, 1] \) for all \( n \in \mathbb{N} \).

---

Would you like a more detailed explanation of any part? Here are some related questions that expand on this topic:

1. How can you find the global maximum of any quadratic function?
2. Can you generalize the recurrence relation for different initial conditions?
3. What is the long-term behavior of the sequence \( u_n \) as \( n \to \infty \)?
4. Could we use other methods (e.g., graphical approaches) to verify the inequality \( x(1 - x) \leq \frac{1}{4} \)?
5. How does the recurrence relation change if we modify the coefficient from 3 to another constant?

Tip: When solving recurrence relations, always try to check boundary and critical points to ensure the correct behavior over an interval.

Soit a ∈ [0; 1]. On définit la suite (u_n) par u_0 = a et, pour tout n ∈ N, u_(n+1) = 3u_n(1 - u_n). Montrer que, pour tout x ∈ [0 ; 1], x(1 - x) ≤ 1/4. Démontrer que, pour tout n ∈ N, u_n ∈ [0; 1].

Explore the step-by-step solution to prove the quadratic inequality x(1 - x) ≤ 1/4 for all x ∈ [0; 1] and demonstrate that the sequence u_n defined by u_(n+1) = 3u_n(1 - u_n) remains within [0; 1]. Suitable for advanced high school mathematics students in Grades 11-12.

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