Iranian Mathematical Olympiad

According to the Lagrange's identity, we have $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$. So we have to prove that:
$$(ac + bd)^2 \geq 9(ad - bc)^2 \iff |ac + bd| \geq 3|ad - bc|$$
First, we check the case where none of the numbers are zero. By dividing both sides by $|ac|$, we have:
$$|1 + \frac{b}{a} \cdot \frac{d}{c}| \geq 3|\frac{d}{c} - \frac{b}{a}|$$
Suppose $a, b, c$, and $d$ have the same sign. In this case:
$$|1 + \frac{b}{a} \cdot \frac{d}{c}| \geq 1$$
It suffices to prove that:
$$|\frac{d}{c} - \frac{b}{a}| \leq \frac{1}{3}$$
Now, if we put the positive numbers in set $A$ and the negative numbers in set $B$, each of them will give us $\lfloor \frac{|A|}{2} \rfloor$ and $\lfloor \frac{|B|}{2} \rfloor$ pairs of numbers that have the same sign. So, in total, we will have 4 pairs of numbers that have the same sign (because one of the sets has an even number of elements and the other has an odd number of elements). For each pair $(a_i, b_i)$, we choose a fraction from among $\frac{a_i}{b_i}$, $\frac{b_i}{a_i}$ that doesn't exceed one. Now we have 4 fractions between zero and one, so according to the pigeonhole principle, the distance between two of them is less than $\frac{1}{3}$, and the result follows.

If one of the numbers, like $a$, is zero, then just before the division, by proof by contradiction we have $b$ is non zero and:
$$|\frac{d}{c}| \geq \frac{1}{3}$$
Similar to the previous case, it is possible to find 3 pairs of the same sign where the assigned fractions are between $\frac{1}{3}$ and 1, and again according to the pigeonhole principle, a difference between two fractions will be less than $\frac{1}{3}$. The result follows.

We shall prove a more strong argument; We choose 8 numbers from these nine numbers and denote them by $x_1, y_1, \dots, x_4, y_4$. We arranged them in the way that $|x_i| \leq |y_i|$ and $x_i, y_i$ have the same sign. Consider four points $u_i = (|x_i|, |y_i|)$ and the angle they make with the origin would be in the interval $[0, \frac{\pi}{4}]$. Thus, the angle between two of them, say $u_i, u_j$ is less that or equal to $\frac{\pi}{12}$. Hence, putting $a = x_i, b = y_i$ and $c = x_j, d = y_j$ it follows that

$$|ac + bd| = |x_i x_j + y_i y_j| = u_i \cdot u_j = |u_i| \cdot |u_j| \cos \theta.$$
That is, $(ac + bd)^2 \ge (a^2 + b^2)(c^2 + d^2)\cos^2 \frac{\pi}{12} = \frac{2+\sqrt{3}}{4}(a^2 + b^2)(c^2 + d^2)$.