2019/20 AHSMC Part 1 Problem 7

We can rewrite the inequality as $2(x-y)^2 + (k-2)(x^2 + y^2) \ge 0$. Putting $x = y = 1$ one obtains $k \ge 2$. On the other hand, if $k \ge 2$ this inequality is true for any $x$, $y$ real numbers. Hence the minimum value of $k$ is $2$. The answer is (C).

Alternative approach: The given inequality is equivalent to $kx^2 - 4xy + ky^2 \ge 0$ for any real numbers $x$, $y$. Putting $x = y = 1$ one obtains $k \ge 2$. On the other hand for $k \ge 2$, $kx^2 - 4xy + ky^2 \ge 2x^2 + 2y^2 - 4xy = 2(x-y)^2 \ge 0$ for any $x, y \in \mathbb{R}$.