Problem 1. Positive numbers $a$ and $b$ are such that $a-b=a / b$. Which is greater, $a+b$ or $a b ?$

(V. Kleptsyn, A. Gusev)
Answer: $a b > a + b$.

Solution 1. Consider the desired inequality

$$
a b \vee a + b
$$

Multiply it by the equality $a / b = a - b$ (left side by left side, right side by right side). When multiplying by a positive number (and $a / b$ is positive), the inequality is preserved. We get

$$
a^{2} \vee a^{2} - b^{2}
$$

From this, it is clear that the left side is greater than the right side.

Solution 2. Consider the desired inequality

$$
a b \vee a + b
$$

Add to it the equality $a / b = a - b$, reducing it to the comparison

$$
a b + a / b \vee 2 a
$$

By moving everything to the left side, we can see that it is equivalent to $\frac{a}{b}(b-1)^{2} \vee 0$. To confirm that the left side is greater, it remains to show that $b \neq 1$.

Indeed, suppose that $b = 1$. Substituting into the original equality gives $a - 1 = a$, a contradiction.

Solution 3. Multiplying the equality $a - b = a / b$ by $b$, we get $a b - b^{2} = a$, from which $a b = a + b^{2}$. Then it is sufficient to compare $a + b^{2}$ and $a + b$, that is, to compare $b$ with 1. Suppose that $b \leqslant 1$. Then $a / b \geqslant a > a - b$, a contradiction. Therefore, $b > 1$, from which $a b = a + b^{2} > a + b$.