[ Algebraic inequalities (miscellaneous).]

Authors: Kiepitsin V.A. Gusev A.S.

Positive numbers $a$ and $b$ are such that $a-b=a / b$. Which is greater, $a+b$ or $a b$?
First solution. Consider the desired inequality

$$
a b \vee a+b \text {. }
$$

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.

Second solution. Consider the desired inequality

$$
a b \vee a+b
$$

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

$$
a b+a / b \vee 2 a \text {. }
$$

By moving everything to the left side, we can see that it is equivalent to $\frac{a}{b}(b-1)^{2} v 0$. To be convinced 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.

Third solution. 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 $b \leq 1$. Then $a / b \geq a>a-b$, a contradiction. Therefore, $b>1$, from which $a b=a+b^{2}>a+b$.