1. **Prove** $\forall x,y \in \mathbb{R}: |x - y| \leq |x| + |y|$. By the triangle inequality for real numbers, we have: $$|x - y| = |x + (-y)| \leq |x| + |-y| = |x| + |y|.$$ This completes the proof. 2. **Prove** $\forall x,y \in \mathbb{R}: ||x| - |y|| \leq |x + y|$. Using the triangle inequality: $$|x| = |(x + y) - y| \leq |x + y| + |y| \implies |x| - |y| \leq |x + y|.$$ Similarly, $$|y| = |(y + x) - x| \leq |y + x| + |x| = |x + y| + |x| \implies |y| - |x| \leq |x + y|.$$ Taking absolute values, $$||x| - |y|| \leq |x + y|.$$ 3. **Prove** $\forall x,y \geq 0: \sqrt{x + y} \leq \sqrt{x} + \sqrt{y}$. Square both sides (valid since all terms are nonnegative): $$x + y \leq (\sqrt{x} + \sqrt{y})^2 = x + y + 2\sqrt{xy}.$$ Subtract $x + y$ from both sides: $$0 \leq 2\sqrt{xy},$$ which is true. Hence, $$\sqrt{x + y} \leq \sqrt{x} + \sqrt{y}.$$ 4. **Prove** $(|x + y| = |x| + |y|) \iff (xy \geq 0)$. - ($\Rightarrow$) If $|x + y| = |x| + |y|$, then $x$ and $y$ have the same sign or one is zero, so $xy \geq 0$. - ($\Leftarrow$) If $xy \geq 0$, then $x$ and $y$ have the same sign or one is zero, so $$|x + y| = |x| + |y|.$$ 5. **Prove** $\forall x,y \in \mathbb{R}: |x| + |y| \leq |x + y| + |x - y|$. Square both sides: $$(|x| + |y|)^2 = x^2 + 2|xy| + y^2,$$ $$(|x + y| + |x - y|)^2 = (x + y)^2 + 2|x + y||x - y| + (x - y)^2 = 2x^2 + 2y^2 + 2|x + y||x - y|.$$ Since $|x + y||x - y| \geq |x^2 - y^2|$, we have $$2x^2 + 2y^2 + 2|x + y||x - y| \geq 2x^2 + 2y^2 + 2|x^2 - y^2| \geq x^2 + 2|xy| + y^2,$$ so $$|x| + |y| \leq |x + y| + |x - y|.$$ 6. **Prove** $\forall x,y \in \mathbb{R}, \forall \varepsilon > 0: xy \leq \frac{x^2}{2\varepsilon} + \frac{\varepsilon}{2} y^2$. Start with the nonnegative square: $$\left( \frac{x}{\sqrt{2\varepsilon}} - \sqrt{\frac{\varepsilon}{2}} y \right)^2 \geq 0,$$ which expands to $$\frac{x^2}{2\varepsilon} - xy + \frac{\varepsilon}{2} y^2 \geq 0,$$ rearranged as $$xy \leq \frac{x^2}{2\varepsilon} + \frac{\varepsilon}{2} y^2.$$ **Final answers:** All statements are proven as requested.