Problem 1346

1. Given the inequality $(x-1)(y-1) \ge 1$, we start by expanding and rearranging it:
$$(x-1)(y-1) \ge 1 \implies xy - x - y + 1 \ge 1 \implies xy \ge x + y$$
2. Next, we consider the reciprocal form of $x$ and $y$. By dividing both sides of $xy \ge x + y$ by $xy$, we get:
$$1 \ge \frac{x}{xy} + \frac{y}{xy} \implies 1 \ge \frac{1}{y} + \frac{1}{x} \implies \frac{1}{x} + \frac{1}{y} \le 1$$
3. Now, we need to prove that for sides $a, b, c$ of an arbitrary triangle, the inequality $a^2x + b^2y > c^2$ holds. We use the Cauchy-Schwarz inequality in the form:
$$(a^2x + b^2y)\left(\frac{1}{x} + \frac{1}{y}\right) \ge (a + b)^2$$
4. Substituting $\frac{1}{x} + \frac{1}{y} \le 1$ into the inequality, we get:
$$(a^2x + b^2y) \cdot 1 \ge (a + b)^2 \implies a^2x + b^2y \ge (a + b)^2$$
5. Since $a, b, c$ are sides of a triangle, by the triangle inequality, we have:
$$a + b > c$$
6. Squaring both sides of the triangle inequality, we get:
$$(a + b)^2 > c^2$$
7. Combining the results from steps 4 and 6, we have:
$$a^2x + b^2y \ge (a + b)^2 > c^2$$
8. Therefore, we conclude:
$$a^2x + b^2y > c^2$$

$\blacksquare$