1. **Problem:** For all positive real numbers $x,y$, is it true that $x + y \leq xy$? **Step 1:** Rewrite inequality: $xy - x - y \geq 0$. **Step 2:** Add 1 to both sides: $xy - x - y + 1 \geq 1$. **Step 3:** Factor left side: $(x-1)(y-1) \geq 1$. **Step 4:** Since $x,y > 0$, $(x-1)(y-1) \geq 1$ means both $x-1$ and $y-1$ are either both greater than or equal to 1 or both less than or equal to -1. **Step 5:** Check if this holds for all $x,y > 0$. For example, $x=y=2$, $(2-1)(2-1)=1 \geq 1$ true; but $x=y=0.5$, $(0.5-1)(0.5-1)=0.25 \not\geq 1$ false. **Conclusion:** Statement 1 is false for all positive $x,y$. 2. **Problem:** For all positive real numbers $x,y$, is it true that $x + y > xy$? **Step 1:** Rewrite inequality: $x + y - xy > 0$. **Step 2:** Add 1 and rearrange: $1 > (x-1)(y-1)$. **Step 3:** For $x,y > 0$, $(x-1)(y-1) < 1$. **Step 4:** Check examples: $x=y=0.5$, $(0.5-1)(0.5-1)=0.25 < 1$ true; $x=y=2$, $(2-1)(2-1)=1 \not< 1$ false. **Conclusion:** Statement 2 is false for all positive $x,y$. 3. **Problem:** For all positive real $y$, does there exist positive real $x$ such that $\cos(x) = y$? **Step 1:** Range of $\cos(x)$ for real $x$ is $[-1,1]$. **Step 2:** For $y \in (0,1]$, there exists $x$ with $\cos(x) = y$. **Step 3:** For $y > 1$, no such $x$ exists. **Step 4:** Since $y \in \mathbb{R}^+$, includes values $>1$, statement is false. 4. **Problem:** For all real $x$, is $x^2 + x + 1 > 0$? **Step 1:** Consider quadratic $f(x) = x^2 + x + 1$. **Step 2:** Discriminant $\Delta = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 < 0$. **Step 3:** Since $\Delta < 0$, $f(x)$ has no real roots and is always positive (leading coefficient positive). **Conclusion:** Statement 4 is true. 5. **Problem:** For all real $x$, does there exist real $y$ such that $x^2 + y - 2xy = 0$? **Step 1:** Rearrange: $y - 2xy = -x^2$. **Step 2:** Factor $y(1 - 2x) = -x^2$. **Step 3:** If $1 - 2x \neq 0$, then $y = \frac{-x^2}{1 - 2x}$ exists. **Step 4:** If $1 - 2x = 0 \Rightarrow x = \frac{1}{2}$, 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 equation at $x=\frac{1}{2}$. **Conclusion:** Statement 5 is false. 6. **Problem:** Does there exist real $y$ such that for all real $x$, $\frac{x}{1 + x^2} \leq y$? **Step 1:** Consider function $f(x) = \frac{x}{1 + x^2}$. **Step 2:** Find maximum of $f(x)$ by setting derivative to zero. **Step 3:** $f'(x) = \frac{(1 + x^2) \cdot 1 - x \cdot 2x}{(1 + x^2)^2} = \frac{1 + x^2 - 2x^2}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2}$. **Step 4:** Set numerator zero: $1 - x^2 = 0 \Rightarrow x = \pm 1$. **Step 5:** Evaluate $f(1) = \frac{1}{1 + 1} = \frac{1}{2}$, $f(-1) = \frac{-1}{1 + 1} = -\frac{1}{2}$. **Step 6:** Maximum value is $\frac{1}{2}$. **Step 7:** Choose $y = \frac{1}{2}$, then $f(x) \leq y$ for all $x$. **Conclusion:** Statement 6 is true. **Final answers:** 1. False 2. False 3. False 4. True 5. False 6. True