2011/12 AHSMC Part 1 Problem 13

Dividing throughout by $y = 1$, we have $(z^2 + 1)^3 > m(z^3 + 1)^2$ where $z = \frac{x}{y}$. This is equivalent to $$(1 - m)z^6 + 3z^4 + (3 - 2m)z^3 + 1 - m > 0$$ for any positive real $z$. Hence it is necessary to have $1 - m \ge 0$, i.e. $m \le 1$. If we take $m = 1$, the inequality $(x^2 + y^2)^3 \ge (x^3 + y^3)^2$ is equivalent to $$x^2 y^2\bigl((x - y)^2 + 2x^2 + 2y^2\bigr) > 0,$$ which is clearly true. The answer is (C).