9.6. Let \(a\) and \(b\) be positive numbers. Find the minimum value of the fraction \(\frac{(a+b)(a+2)(b+2)}{16ab}\). Justify your answer.
Solution: By the inequality between the arithmetic mean and the geometric mean of two numbers, the following three inequalities hold:

$$
a+b \geqslant 2 \sqrt{a b}, \quad a+2 \geqslant 2 \sqrt{2 a}, \quad b+2 \geqslant 2 \sqrt{b}
$$

Multiplying the left and right sides of these three inequalities, we get

$$
(a+b)(a+2)(b+2) \geqslant 8 \sqrt{a b \cdot 2 a \cdot 2 b}
$$

which is equivalent to the inequality $\frac{(a+b)(a+2)(b+2)}{16 a b} \geqslant 1$. The value 1 is achieved at the set $a=b=2$.

Note: The fraction given in the problem is a special case of the fraction $\frac{(a+b)(a+c)(b+c)}{8 a b c}$. Its minimum (for positive variables) is also 1 and is achieved when $a=b=c$. The proof of this fact does not differ from the one provided in the solution.

## Answer: 1.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| The equation $\frac{(a+b)(a+2)(b+2)}{16 a b}=t$ (or an equivalent one) is investigated as a quadratic (in one of the variables) with parameters, but the investigation is incomplete or incorrect | 4 points |
| The case $a=b$ is correctly investigated, but it is not justified that the minimum of the fraction is achieved precisely under this assumption | 3 points |
| The set at which the minimum value of the fraction is achieved $(a=b=2)$ is indicated, but its optimality is not proven | 1 point |
| Correct answer without justification and/or any calculations from which the solution process is not clear | 0 points |