# Solution:

Since the numbers $a, b, c, d$ are positive, multiplying both sides of the inequality by the product of the denominators yields an equivalent inequality:

$$(b+d)\left(d a^{2}+b c^{2}\right) \geq b d(a+c)^{2}$$

Expanding the brackets: $a^{2} b d+a^{2} d^{2}+b^{2} c^{2}+b c^{2} d \geq a^{2} b d+2 a b c d+b c^{2} d$.

Moving all terms to the left side of the inequality and combining like terms: $a^{2} d^{2}-2 a b c d+b^{2} c^{2} \geq 0$

The last inequality can be written as: $(a d-b c)^{2} \geq 0$.

Since we obtained a true inequality through equivalent transformations, the original inequality is also true.

## Criteria:

If the positivity of the numbers is not mentioned when multiplying by the common denominator, deduct 2 points.