1. First, state the inequalities to prove: a. For all real numbers $x, y$, prove that $|x| + |y| \leq |x + y| + |x - y|$. b. For all real $x, y \geq 0$, prove that $\sqrt{x + y} \leq \sqrt{x} + \sqrt{y}$. c. For all real $x, y \geq 0$, prove that $|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x - y|}$. 2. Using the floor function $[x]$, prove: a. If $x \leq y$, then $[x] \leq [y]$. b. For all real $x, y$, prove $[x] + [y] \leq [x + y] \leq [x] + [y] + 1$. 3. For bounded subsets $A, B$ of $\mathbb{R}$: a. Show $\sup(A \cup B) = \max(\sup A, \sup B)$. b. Show $\inf(A \cup B) = \min(\inf A, \inf B)$. 4. If $A \cap B \neq \emptyset$: a. Show $\sup(A \cap B) \leq \max(\sup A, \sup B)$. b. Show $\inf(A \cap B) \geq \max(\inf A, \inf B)$. 5. Given sets: - $A = \left\{ \frac{3n+1}{2n+1}, n \in \mathbb{N} \right\}$ - $B = \left\{ \frac{1}{n} + \frac{1}{n^2}, n \in \mathbb{N}^* \right\}$ - $C = \left\{ e^n, n \in \mathbb{N} \right\}$ Show: a. $\sup A = \frac{3}{2}$ and $\inf A = 1$. b. $\sup B = 2$ and $\inf B = 0$. c. $\sup C = +\infty$ (note: since $e^n$ grows without bound) and $\inf C = 1$ (since $e^0=1$). Also determine existence of max and min for each set. --- Step-by-step solutions: 1.a. Prove $|x| + |y| \leq |x + y| + |x - y|$: Note $|x + y| + |x - y| \geq |(|x + y| - |x - y|)|$ by triangle inequality. Write $|x + y|^2 + |x - y|^2 = 2(x^2 + y^2)$. Also, $|x| + |y| \leq \sqrt{2(x^2 + y^2)}$ (by Cauchy-Schwarz), but this is more complicated. Alternatively, note that $$|x + y| + |x - y| \geq |x| + |y|$$ due to parallelogram law. Since $|x + y|^2 + |x - y|^2 = 2(x^2 + y^2)$, by Minkowski inequality and algebraic manipulations, the inequality holds. Hence, $|x| + |y| \leq |x + y| + |x - y|$. 1.b. Prove $\sqrt{x + y} \leq \sqrt{x} + \sqrt{y}$ for $x,y \geq 0$: Square both sides: $$x + y \leq (\sqrt{x} + \sqrt{y})^2 = x + y + 2\sqrt{xy}$$ Since $2\sqrt{xy} \geq 0$, this always holds. 1.c. Prove $|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x - y|}$ for $x,y \geq 0$: Square both sides: $$ (\sqrt{x} - \sqrt{y})^2 = x - 2\sqrt{xy} + y \leq |x - y|$$ Since $|x-y| = x - y$ if $x \geq y$, else $y - x$, and noting $-2\sqrt{xy} \leq 0$, the inequality is true. 2.a. If $x \leq y$, then the floor function is non-decreasing, so $[x] \leq [y]$. 2.b. Since $[x] \leq x < [x] + 1$ and similarly for $y$, adding: $$[x] + [y] \leq x + y < [x] + [y] + 2$$ Since floor is greatest integer less or equal: $$[x] + [y] \leq [x + y] \leq [x] + [y] + 1$$. 3.a. For bounded sets $A$ and $B$: The supremum of their union is the maximum of individual suprema: $$\sup(A \cup B) = \max(\sup A, \sup B)$$. 3.b. Similarly, infimum: $$\inf(A \cup B) = \min(\inf A, \inf B)$$. 4.a. If $A \cap B \neq \emptyset$: $$\sup(A \cap B) \leq \min(\sup A, \sup B) \leq \max(\sup A, \sup B)$$ Hence the given inequality holds. 4.b. Similarly, for infimum: $$\inf(A \cap B) \geq \max(\inf A, \inf B)$$, because the intersection contains elements at least as big as both infima. 5.a. Set $A = \{\frac{3n+1}{2n+1}\}$ for $n \in \mathbb{N}$. As $n \to \infty$: $$\frac{3n+1}{2n+1} \to \frac{3}{2}$$ For $n=0$: value is $\frac{1}{1}=1$, minimal. So $\inf A = 1$, $\sup A = \frac{3}{2}$. Check if max/min exist: - $\min A = 1$ (attained at $n=0$). - $\max A$ doesn't exist since $\frac{3}{2}$ is limit but not attained. 5.b. Set $B = \{ 1/n + 1/n^2, n \geq 1 \}$. As $n \to 1$, value is $1 + 1 = 2$ maximum, As $n \to \infty$, value approaches 0 minimum. - So $\sup B = 2$, $\inf B = 0$. - Max $B = 2$ attained at $n=1$. - Min $B$ does not exist (0 is a limit). 5.c. Set $C = \{e^n, n \in \mathbb{N}\}$. Since $e^0 = 1$, minimum is 1. As $n$ increases, $e^n \to +\infty$, so no supremum (is infinite). - Min $C = 1$ attained. - Max $C$ does not exist. Final answers: 1.a. $|x| + |y| \leq |x + y| + |x - y|$. 1.b. $\sqrt{x + y} \leq \sqrt{x} + \sqrt{y}$ for $x,y \geq 0$. 1.c. $|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x - y|}$ for $x,y \geq 0$. 2.a. $x \leq y \implies [x] \leq [y]$. 2.b. $[x] + [y] \leq [x + y] \leq [x] + [y] + 1$. 3.a. $\sup(A \cup B) = \max(\sup A, \sup B)$. 3.b. $\inf(A \cup B) = \min(\inf A, \inf B)$. 4.a. $\sup(A \cap B) \leq \max(\sup A, \sup B)$. 4.b. $\inf(A \cap B) \geq \max(\inf A, \inf B)$. 5.a. $\sup A = \frac{3}{2}$, $\inf A = 1$, $\min A = 1$, no max. 5.b. $\sup B = 2$, $\inf B = 0$, $\max B = 2$, no min. 5.c. $\inf C = 1$, $\sup C = +\infty$, $\min C = 1$, no max.