Problem 1636

1. We start with the given inequality:
$$\frac{60a^2 - 1}{4xy + 5z} + \frac{60b^2 - 1}{4yz + 5x} + \frac{60c^2 - 1}{4zx + 5y} \geq 12$$
where $a, b, c$ are permutations of $x, y, z$ and $x, y, z \in \left[\frac{1}{2}, 2\right]$.

2. Consider the inequalities:
$$(x - 2)(y - \frac{1}{2}) \leq 0 \quad \text{and} \quad (x - \frac{1}{2})(y - 2) \leq 0$$
These inequalities are equivalent to:
$$xy - 2y - \frac{1}{2}x + 1 \leq 0 \quad \text{and} \quad xy - 2x - \frac{1}{2}y + 1 \leq 0$$

3. Adding these two inequalities, we get:
$$2xy - 2y - \frac{1}{2}x - 2x - \frac{1}{2}y + 2 \leq 0$$
Simplifying, we obtain:
$$2xy \leq \frac{5}{2}(x + y) - 2$$

4. Multiplying by 2 and adding $5z$ to both sides, we get:
$$4xy + 5z \leq 5(x + y + z) - 4$$

5. Similarly, we can derive:
$$4yz + 5x \leq 5(x + y + z) - 4 \quad \text{and} \quad 4zx + 5y \leq 5(x + y + z) - 4$$

6. Substituting these into the original inequality, we have:
$$\frac{60a^2 - 1}{5(x + y + z) - 4} + \frac{60b^2 - 1}{5(x + y + z) - 4} + \frac{60c^2 - 1}{5(x + y + z) - 4} \geq 12$$

7. Combining the fractions, we get:
$$\frac{60(a^2 + b^2 + c^2) - 3}{5(x + y + z) - 4} \geq 12$$

8. Multiplying both sides by $5(x + y + z) - 4$, we obtain:
$$60(a^2 + b^2 + c^2) - 3 \geq 12 \cdot (5(x + y + z) - 4)$$

9. Simplifying the right-hand side, we get:
$$60(a^2 + b^2 + c^2) - 3 \geq 60(x + y + z) - 48$$

10. Rearranging terms, we have:
$$60(a^2 + b^2 + c^2) - 60(x + y + z) \geq -45$$

11. Dividing by 15, we get:
$$4(a^2 + b^2 + c^2) - 4(x + y + z) \geq -3$$

12. Adding 3 to both sides, we obtain:
$$4(a^2 + b^2 + c^2) + 3 \geq 4(x + y + z)$$

13. Grouping terms to form squares, we get:
$$(2a - 1)^2 + (2b - 1)^2 + (2c - 1)^2 \geq 0$$

Since the sum of squares is always non-negative, the inequality holds true.

$\blacksquare$