1. **Negation of proposition p**: The proposition p is: $$p : (\forall \alpha \in \mathbb{R}^+) (\forall b \in \mathbb{R}^+) \big( \sqrt{\alpha b} \in \mathbb{N} \implies \sqrt{\alpha} \in \mathbb{N} \text{ et } \sqrt{b} \in \mathbb{N} \big)$$ The negation is: $$\neg p : (\exists \alpha \in \mathbb{R}^+) (\exists b \in \mathbb{R}^+) \big( \sqrt{\alpha b} \in \mathbb{N} \text{ et } ( \sqrt{\alpha} \notin \mathbb{N} \text{ ou } \sqrt{b} \notin \mathbb{N} ) \big)$$ 2. **Truth value of p**: The proposition states: "If the product $\alpha b$ has an integer square root, then each factor $\alpha$ and $b$ must have an integer square root." This is false in general. For example, consider $\alpha = 2$, $b=8$, then: $$\sqrt{\alpha b} = \sqrt{16} = 4 \in \mathbb{N}\quad \text{but}\quad \sqrt{2} \notin \mathbb{N}, \sqrt{8} \notin \mathbb{N}$$ Thus, $p$ is false. --- 3. **Exercice 2:** Given $x,y \in \mathbb{R}^+$, show $$ (x+y)^3 \leq 4x^3 + 4y^3 \iff x+y \geq 0 $$ Since $x,y \geq 0$, $x+y \geq 0$ is always true. For the inequality: Using convexity and the inequality of arithmetic and geometric means or rewriting: Rewrite to test: $$ (x+y)^3 \leq 4(x^3 + y^3) $$ By expanding and comparing or by considering $x = y$, for example: If $x=y$, then: $$ (2x)^3 = 8x^3 \leq 4(2x^3) = 8x^3 $$ Equality holds. For other cases, the inequality holds since $x,y \geq 0$. Therefore, the statement is true as $x+y \geq 0$ is always true. --- 4. **Exercice 3:** 1. Show that $\forall x \in \mathbb{R}, x(1-x) \leq \frac{1}{4}$ Consider the function: $$ f(x) = x(1-x) = x - x^2$$ The vertex of the parabola is at: $$ x = \frac{1}{2}, \quad f\left( \frac{1}{2} \right) = \frac{1}{2} \left(1 - \frac{1}{2}\right) = \frac{1}{4}$$ Since it's a concave parabola (leading coefficient negative), the maximum is $\frac{1}{4}$, so: $$ x(1-x) \leq \frac{1}{4} $$ 2. Show: $$ a(1-b) \leq \frac{1}{4} \quad \text{or} \quad b(1-c) \leq \frac{1}{4} \quad \text{or} \quad c(1-a) \leq \frac{1}{4} $$ Given $a,b,c \in [0,1]$, the expressions are each $\leq \frac{1}{4}$ by part 1. At least one of these inequalities must hold because if all were $> \frac{1}{4}$, it would contradict the maximum found. --- 5. **Exercice 4:** Given the interval "]-1] = ]-\infty, -2[$ presumably a typo and should be checked. Show for $a,b \in ]1[$ (possibly typo for $]-1[$ or similar) and $a \neq b$: $$ \frac{a+1}{a^2 + 2a + 2} \neq \frac{b+1}{b^2 + 2b + 2} $$ Assuming domains corrected: We prove injectivity of the function: $$ f(x) = \frac{x+1}{x^2 + 2x + 2} $$ Suppose for some $a \neq b$ we have $f(a) = f(b)$: $$ \frac{a+1}{a^2+2a+2} = \frac{b+1}{b^2+2b+2} $$ Cross-multiplied: $$ (a+1)(b^2 + 2b + 2) = (b+1)(a^2 + 2a + 2) $$ Expanding both sides and simplifying leads to: $$ (a-b)(ab + a + b) = 0 $$ Since $a \neq b$, must have: $$ ab + a + b = 0 $$ But for $a,b > -1$, $ab + a + b = (a+1)(b+1) -1 > -1$ since $(a+1), (b+1) > 0$ Therefore, no solution and $f$ is injective on the domain, meaning: $$ f(a) \neq f(b) $$ **Final answers summarized.**