1. Problem 1: Prove or disprove the inequality $x^3 + y^3 < (x + y)^3$. 2. The student expands $(x + y)^3$ correctly: $$ (x + y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3 $$ 3. The student claims $(x + y)^3 < x^3 + y^3$ because $3x^2 y + 3xy^2 > 0$. 4. Error: The student incorrectly assumes $3x^2 y + 3xy^2 > 0$ implies $(x + y)^3 < x^3 + y^3$. Actually, since $3x^2 y + 3xy^2$ is added to $x^3 + y^3$, the correct inequality is: $$ (x + y)^3 = x^3 + y^3 + 3xy(x + y) $$ which is greater than or equal to $x^3 + y^3$ if $3xy(x + y) \\geq 0$. 5. Counter-example: Let $x = -1$, $y = 1$. Calculate: $$ x^3 + y^3 = (-1)^3 + 1^3 = -1 + 1 = 0 $$ $$ (x + y)^3 = (0)^3 = 0 $$ Here, $x^3 + y^3 = (x + y)^3$, so the strict inequality $x^3 + y^3 < (x + y)^3$ is false. 6. Problem 2: Verify the inequality $x + y \\geq \sqrt{x^2 + y^2}$ for negative $x,y$. 7. The student squares both sides: $$ (x + y)^2 \\geq x^2 + y^2 $$ which expands to: $$ x^2 + 2xy + y^2 \\geq x^2 + y^2 $$ leading to: $$ 2xy \\geq 0 $$ 8. Error: The student claims $2xy > 0$ because $x,y$ are negative, so $xy$ is positive. This is correct, but squaring both sides can introduce extraneous solutions and the original inequality may not hold. 9. Counter-example: Let $x = y = -1$. Calculate: $$ x + y = -1 -1 = -2 $$ $$ \sqrt{x^2 + y^2} = \sqrt{1 + 1} = \sqrt{2} \approx 1.414 $$ Since $-2 \\not\geq 1.414$, the inequality fails. 10. Proof for positive $x,y$: For $x,y > 0$, by Cauchy-Schwarz inequality or by direct comparison: $$ x + y \\geq \sqrt{x^2 + y^2} $$ Square both sides (valid since both sides positive): $$ (x + y)^2 = x^2 + 2xy + y^2 \\geq x^2 + y^2 $$ Since $2xy \\geq 0$, the inequality holds. Final answers: - Error in problem 1: Incorrect sign assumption and inequality direction. - Counter-example: $x=-1,y=1$. - Error in problem 2: Squaring both sides can introduce extraneous solutions; original inequality fails for negative $x,y$. - Counter-example: $x=y=-1$. - Inequality true for positive $x,y$ by squaring and positivity.