Question

(a+b)(a+2)(b+2)>=16ab Докажите неравенство. Подробно пожалуйста

Answer 1

Пусть a=5, а b=10, тогда:
(5+2)(10+2)(5+10)>=16*5*10
1260>=800

Answer 2

To prove the inequality (a+b)(a+2)(b+2) >= 16ab, we can expand both sides of the inequality and simplify the expression.

Expanding the left side of the inequality: (a+b)(a+2)(b+2) = (a^2 + 2a + ab + 2b)(b+2) = a^2b + 2ab + ab^2 + 2a^2 + 4a + 2b + 4ab + 8 = a^2b + 6ab + ab^2 + 2a^2 + 4a + 2b + 8

Now, let's expand the right side of the inequality: 16ab

Comparing the expanded expressions, we have: a^2b + 6ab + ab^2 + 2a^2 + 4a + 2b + 8 >= 16ab

To prove the inequality, we need to show that the left side is greater than or equal to the right side for all possible values of a and b.

Let's simplify the expression further: a^2b + 6ab + ab^2 + 2a^2 + 4a + 2b + 8 - 16ab >= 0 a^2b + ab^2 + 2a^2 - 10ab + 4a + 2b + 8 >= 0

Now, let's factor out common terms: ab(a + b) + 2a(a - 5b) + 2(b + 4) >= 0

To prove the inequality, we need to show that the left side is greater than or equal to zero for all possible values of a and b.

Since the inequality involves multiple variables, we can't directly prove it by substituting values. Instead, we can analyze different cases to determine when the inequality holds true.

Case 1: a + b > 0 If a + b > 0, then ab(a + b) >= 0. So, we need to prove that 2a(a - 5b) + 2(b + 4) >= 0.

Case 1.1: a - 5b >= 0 If a - 5b >= 0, then 2a(a - 5b) >= 0. So, we need to prove that 2(b + 4) >= 0.

Case 1.1.1: b + 4 >= 0 If b + 4 >= 0, then 2(b + 4) >= 0. Therefore, the inequality holds true for this case.

Case 1.1.2: b + 4 < 0 If b + 4 < 0, then 2(b + 4) < 0. Therefore, the inequality does not hold true for this case.

Case 1.2: a - 5b < 0 If a - 5b < 0, then 2a(a - 5b) < 0. Therefore, the inequality does not hold true for this case.

Case 2: a + b < 0 If a + b < 0, then ab(a + b) <= 0. So, we need to prove that 2a(a - 5b) + 2(b + 4) >= 0.

Case 2.1: a - 5b >= 0 If a - 5b >= 0, then 2a(a - 5b) >= 0. So, we need to prove that 2(b + 4) >= 0.

Case 2.1.1: b + 4 >= 0 If b + 4 >= 0, then 2(b + 4) >= 0. Therefore, the inequality holds true for this case.

Case 2.1.2: b + 4 < 0 If b + 4 < 0, then 2(b + 4) < 0. Therefore, the inequality does not hold true for this case.

Case 2.2: a - 5b < 0 If a - 5b < 0, then 2a(a - 5b) < 0. Therefore, the inequality does not hold true for this case.

Based on the analysis of different cases, we can conclude that the inequality (a+b)(a+2)(b+2) >= 16ab does not hold true for all possible values of a and b.

Please note that this is a proof by case analysis, and it demonstrates that the inequality does not hold true for all possible values of a and b.