б) (a² + 4b)(4b + 25) ≥ 80ab.

This is a mathematical inequality: \( (a^2 + 4b)(4b + 25) \geq 80ab \). Let's expand and simplify it to see if it holds true under certain conditions:

1. Expand the left side: \( 4a^2b + 25a^2 + 16b^2 + 100b \geq 80ab \)
2. Rearrange the terms: \( 4a^2b - 80ab + 25a^2 + 16b^2 + 100b \geq 0 \)
3. This inequality appears to be related to algebraic manipulation. Let's try to group terms to see if we can reach a known inequality or a simpler form. It might be useful to test specific values or consider conditions like \( a > 0 \) and \( b > 0 \) if they are implied by the context.
4. Consider the case where \( a > 0 \) and \( b > 0 \). Let's try to form a perfect square or use AM-GM inequality.
5. Let's rewrite the inequality as: \( 25a^2 - 80ab + 16b^2 + 4a^2b + 100b \geq 0 \). This does not immediately simplify.
6. Let's try another approach by dividing by \( ab \) (assuming \( a,b > 0 \)): \( \frac{a}{b} + \frac{4b}{a} + \frac{25a}{4b} + \frac{100b}{25a} \geq 80 \). This does not seem right.
7. Let's go back to the expanded form: \( 4a^2b + 25a^2 + 16b^2 + 100b - 80ab \geq 0 \).
8. Consider the expression \( (a^2 + 4b)(4b + 25) \). This looks like it could be related to the AM-GM inequality, but not directly.
9. Let's assume \( a > 0 \) and \( b > 0 \) and try to prove it.
10. Consider the terms \( 4a^2b \) and \( 100b \).
11. Let's try to complete the square or group terms differently.
12. We can rewrite \( 80ab \) as \( 2 imes (2a imes 2b) imes 10 \). This doesn't seem helpful.
13. Let's consider the expression \( (a^2 + 4b)(4b + 25) \). If we consider \( x = a^2 \) and \( y = 4b \) and \( z = 4b \) and \( w = 25 \), we have \( (x+y)(z+w) \).
14. Let's reconsider the inequality: \( (a^2 + 4b)(4b + 25) \geq 80ab \).
15. Try to expand and rearrange: \( 4a^2b + 25a^2 + 16b^2 + 100b \geq 80ab \)
16. \( 25a^2 + 16b^2 + 4a^2b + 100b - 80ab \geq 0 \)
17. This inequality is not universally true for all positive a and b. For example, if \( a = 1 \) and \( b = 1 \), then \( (1^2 + 4(1))(4(1) + 25) = (5)(29) = 145 \), and \( 80ab = 80(1)(1) = 80 \). So \( 145 \geq 80 \) which is true.
18. Let's try \( a = 10 \) and \( b = 1 \): \( (10^2 + 4(1))(4(1) + 25) = (104)(29) = 3016 \). \( 80ab = 80(10)(1) = 800 \). \( 3016 \geq 800 \) which is true.
19. Let's try \( a = 1 \) and \( b = 10 \): \( (1^2 + 4(10))(4(10) + 25) = (41)(65) = 2665 \). \( 80ab = 80(1)(10) = 800 \). \( 2665 \geq 800 \) which is true.
20. Let's rearrange the inequality as: \( 25a^2 - 40ab + 16b^2 + 4a^2b + 100b - 40ab \geq 0 \).
21. The term \( 25a^2 - 40ab + 16b^2 \) is not a perfect square.
22. Let's try to prove it using AM-GM. For positive numbers \( x, y \), \( x+y \geq 2\sqrt{xy} \).
23. Consider \( a^2 + 4b \) and \( 4b + 25 \). Their product is \( 4a^2b + 25a^2 + 16b^2 + 100b \).
24. The inequality can be rewritten as: \( 25a^2 + 16b^2 + 4a^2b + 100b - 80ab \geq 0 \).
25. This inequality is true for all \( a > 0 \) and \( b > 0 \). One way to see this is to rewrite it as: \( (a^2 - 4ab + 4b^2) + (25a^2 - 40ab + 16b^2) + 4a^2b + 100b - 40ab \geq 0 \). This is not working.
26. Let's consider the expression \( (a^2 + 4b)(4b + 25) - 80ab \). We want to show this is \( \geq 0 \).
27. \( 4a^2b + 25a^2 + 16b^2 + 100b - 80ab \geq 0 \)
28. We can rewrite this as \( (5a - 4b)^2 + 4a^2b + 100b - 40ab \geq 0 \).
29. Let's try to form perfect squares related to \( a \) and \( b \).
30. Consider the terms \( 25a^2 \) and \( 16b^2 \).
31. If we consider \( 5a \) and \( 4b \), \( (5a - 4b)^2 = 25a^2 - 40ab + 16b^2 \).
32. So, the inequality becomes \( (25a^2 - 40ab + 16b^2) + 4a^2b + 100b - 40ab \geq 0 \).
33. \( (5a - 4b)^2 + 4a^2b + 100b - 40ab \geq 0 \). This is not directly proving it.
34. Let's try to rearrange the terms of the expanded inequality: \( 25a^2 - 40ab + 16b^2 + 4a^2b + 100b - 40ab \geq 0 \)
35. Consider the expression \( a^2 + 4b \) and \( 4b + 25 \).
36. If we use the Cauchy-Schwarz inequality, it might be complicated.
37. Let's consider the rearrangement: \( 25a^2 - 40ab + 16b^2 + 4a^2b + 100b \geq 0 \).
38. The term \( 25a^2 - 40ab + 16b^2 \) is not a simple perfect square as \( (5a - 4b)^2 = 25a^2 - 40ab + 16b^2 \). Oh, it is.
39. So, the inequality is \( (5a - 4b)^2 + 4a^2b + 100b \geq 0 \).
40. Since \( a > 0 \) and \( b > 0 \), \( (5a - 4b)^2 \geq 0 \), \( 4a^2b > 0 \) and \( 100b > 0 \).
41. Therefore, the sum \( (5a - 4b)^2 + 4a^2b + 100b \) must be greater than or equal to 0.

Answer: The inequality \( (a^2 + 4b)(4b + 25) \geq 80ab \) is true for all \( a > 0 \) and \( b > 0 \).