Product bound from sum of squares

Let the two numbers be $a$ and $b$. Given $a^2 + b^2 = 97$.

We know that $(a - b)^2 \ge 0 \implies a^2 + b^2 \ge 2ab$. $97 \ge 2ab \implies ab \le 48.5$

Also $(a + b)^2 \ge 0 \implies a^2 + b^2 \ge -2ab$. $97 \ge -2ab \implies ab \ge -48.5$

Thus, $-48.5 \le ab \le 48.5$. Among the options, 64 is greater than 48.5 and hence cannot be their product.