1. **Problem:** Determine the truth of the statement $\forall x \in \mathbb{R}^+, \forall y \in \mathbb{R}^+ : x + y \leq xy$. **Step 1:** Recall that for positive real numbers, the inequality $x + y \leq xy$ can be rearranged as $xy - x - y \geq 0$. **Step 2:** Add 1 to both sides: $xy - x - y + 1 \geq 1$. **Step 3:** Factor the left side: $(x - 1)(y - 1) \geq 1$. **Step 4:** Since $x,y > 0$, if both $x,y > 1$, then $(x-1)(y-1) > 0$, but is it always $\geq 1$? For example, $x=1.5$, $y=1.5$ gives $(0.5)(0.5)=0.25 < 1$. **Conclusion:** The statement is false. 2. **Problem:** Determine the truth of $\forall x \in \mathbb{R}^+, \forall y \in \mathbb{R}^+ : x + y \geq xy$. **Step 1:** Using the same factorization, $x + y \geq xy$ implies $(x-1)(y-1) \leq 1$. **Step 2:** For $x,y$ close to 1, $(x-1)(y-1)$ is small, so inequality holds. **Step 3:** For $x,y$ large, e.g. $x=4,y=5$, $(3)(4)=12 > 1$, so $x + y = 9 < 20 = xy$, inequality fails. **Conclusion:** The statement is false. 3. **Problem:** Is $\forall y \in \mathbb{R}^+, \exists x \in \mathbb{R}^+ : \cos(x) = y$ true? **Step 1:** The cosine function ranges between $-1$ and $1$. **Step 2:** Since $y > 0$, $y$ is in $(0, \infty)$. **Step 3:** But cosine only attains values in $[-1,1]$, so for $y > 1$, no $x$ satisfies $\cos(x) = y$. **Conclusion:** The statement is false. 4. **Problem:** Show $\forall x \in \mathbb{R} : x^2 + x + 1 > 0$. **Step 1:** Consider the quadratic $f(x) = x^2 + x + 1$. **Step 2:** Compute discriminant $\Delta = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 < 0$. **Step 3:** Since $\Delta < 0$ and leading coefficient positive, $f(x) > 0$ for all real $x$. **Conclusion:** The statement is true. 5. **Problem:** For $P : \forall x \in \mathbb{R}, \exists y \in \mathbb{R} : x^2 + y - 2xy = 0$, determine if true. **Step 1:** Rearrange to solve for $y$: $y - 2xy = -x^2$. **Step 2:** Factor $y(1 - 2x) = -x^2$. **Step 3:** If $1 - 2x \neq 0$, then $y = \frac{-x^2}{1 - 2x}$. **Step 4:** If $1 - 2x = 0$, i.e. $x = \frac{1}{2}$, then equation is $\left(\frac{1}{2}\right)^2 + y - 2 \cdot \frac{1}{2} \cdot y = \frac{1}{4} + y - y = \frac{1}{4} \neq 0$. **Step 5:** No $y$ satisfies the equation at $x=\frac{1}{2}$. **Conclusion:** The statement $P$ is false. 6. **Problem:** For $Q : \exists y \in \mathbb{R}, \forall x \in \mathbb{R} : \frac{x}{1 + x^2} \leq y$, determine if true. **Step 1:** Consider $f(x) = \frac{x}{1 + x^2}$. **Step 2:** Find the supremum of $f(x)$ over $\mathbb{R}$. **Step 3:** Compute derivative: $$f'(x) = \frac{(1 + x^2) \cdot 1 - x \cdot 2x}{(1 + x^2)^2} = \frac{1 + x^2 - 2x^2}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2}.$$ **Step 4:** Set $f'(x) = 0$ gives $x = \pm 1$. **Step 5:** Evaluate $f(1) = \frac{1}{2} = 0.5$, $f(-1) = \frac{-1}{2} = -0.5$. **Step 6:** As $x \to \pm \infty$, $f(x) \to 0$. **Step 7:** Maximum value is $0.5$, so choose $y = 0.5$. **Conclusion:** The statement $Q$ is true. **Summary:** - 1: False - 2: False - 3: False - 4: True - 5: False - 6: True **Desmos graph:** Triangle with vertices at $(1,1)$, $(4,1)$, $(4,5)$. **Desmos latex:** $y=\begin{cases}1 & 1 \leq x \leq 4 \\ \frac{4}{3}x - \frac{1}{3} & 1 \leq x \leq 4 \end{cases}$ (approximate lines forming triangle edges).