Kvant Math Problem 1281

Consider two positive numbers $x$ and $y$ such that $xy > x + y$.

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Problem

Prove that if the product of two positive numbers is greater than their sum, then their sum is greater than 4.

N. B. Vasiliev

Tournament of Towns (Autumn 1990)

Exploration

Consider two positive numbers $x$ and $y$ such that $xy > x + y$. Trying small integer pairs shows that the inequality fails for numbers less than or equal to $2$. For example, $x = y = 1$ gives $xy = 1$ and $x + y = 2$, so $1 \not> 2$. $x = y = 2$ gives $xy = 4$ and $x + y = 4$, so $4 \not> 4$. $x = y = 3$ gives $xy = 9$ and $x + y = 6$, so $9 > 6$, which works. This suggests that if $xy > x + y$, then both $x$ and $y$ are strictly greater than $2$. Testing slightly asymmetric cases, such as $x = 1.5$ and $y = 3$, yields $xy = 4.5$ and $x + y = 4.5$, which is equality, so still not strictly greater. This exploration indicates that the critical boundary is near $2$, and the sum must exceed $4$.

The core insight is that the inequality $xy > x + y$ is equivalent to $(x - 1)(y - 1) > 1$, which transforms the problem into a simpler form where the lower bounds on $x + y$ become transparent. The most delicate step is justifying that $(x - 1)(y - 1) > 1$ indeed implies $x + y > 4$ for all positive $x$ and $y$.

Problem Understanding

We are asked to prove a universal statement: if the product of two positive numbers exceeds their sum, then their sum exceeds $4$. This is a Type B problem. The difficulty lies in transforming the inequality $xy > x + y$ into a form that allows direct reasoning about the sum $x + y$. The crucial observation is that subtracting $x + y$ from both sides and adding $1$ gives a factorization that yields a lower bound on $x + y$. The problem is elementary in appearance but requires precise manipulation of inequalities and careful handling of positivity.

Proof Architecture

Lemma 1: For any positive numbers $x$ and $y$, the inequality $xy > x + y$ is equivalent to $(x - 1)(y - 1) > 1$. This follows by expanding $(x - 1)(y - 1)$ and rearranging terms.

Lemma 2: If $x - 1 > 0$ and $y - 1 > 0$, and $(x - 1)(y - 1) > 1$, then $x + y = (x - 1) + (y - 1) + 2 > 4$. This follows from AM-GM applied to $(x - 1)$ and $(y - 1)$, but it also can be verified directly since the sum exceeds twice the square root of the product, which is greater than $2$.

The hardest step is ensuring that both $x - 1$ and $y - 1$ are positive, i.e., $x > 1$ and $y > 1$, which must be justified from the initial inequality $xy > x + y$. This requires considering cases where one variable is $\le 1$ and showing these are impossible.

Solution

Let $x$ and $y$ be positive numbers satisfying $xy > x + y$. Consider first the possibility that either $x \le 1$ or $y \le 1$. Without loss of generality, assume $x \le 1$. Then $xy \le y$, since $x \le 1$, and $x + y \ge y + 0 = y$. Therefore $xy \le y \le x + y$, contradicting $xy > x + y$. This contradiction shows that both $x$ and $y$ must be strictly greater than $1$, so $x - 1 > 0$ and $y - 1 > 0$.

Rewrite the inequality as $xy - x - y > 0$ and add $1$ to both sides to obtain

$(x - 1)(y - 1) = xy - x - y + 1 > 1.$

Since $x - 1 > 0$ and $y - 1 > 0$, we can consider their sum:

$x + y = (x - 1) + (y - 1) + 2.$

The inequality $(x - 1)(y - 1) > 1$ implies that $(x - 1) + (y - 1) > 2$ because for any two positive numbers $a$ and $b$, if $ab > 1$, then $a + b > 2$, which follows from the AM-GM inequality applied to $a$ and $b$:

$\frac{a + b}{2} \ge \sqrt{ab} > 1 \implies a + b > 2.$

Substituting $a = x - 1$ and $b = y - 1$, we obtain

$(x - 1) + (y - 1) > 2 \implies x + y = (x - 1) + (y - 1) + 2 > 4.$

This completes the proof.

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Verification of Key Steps

The first delicate step is showing $x > 1$ and $y > 1$. Testing $x = 1$ or $y = 1$ yields equality $xy = x + y$, so no counterexample arises. If $x < 1$ and $y > 1$, then $xy < y < x + y$, confirming the impossibility. Testing $x, y \in (0,1)$ gives $xy < x + y$, also impossible. Therefore the deduction that $x, y > 1$ is rigorous.

The second delicate step is applying AM-GM to $(x - 1)$ and $(y - 1)$. Suppose $(x - 1)(y - 1) = 1 + \epsilon$ with $\epsilon > 0$. Then $\sqrt{(x - 1)(y - 1)} = \sqrt{1 + \epsilon} > 1$, so $(x - 1) + (y - 1) \ge 2\sqrt{(x - 1)(y - 1)} > 2$, confirming $x + y > 4$. Direct substitution of numeric examples, such as $x = 2.5$, $y = 2$, gives $xy = 5 > 4.5 = x + y$ and $x + y = 4.5 > 4$, consistent with the inequality.

Alternative Approaches

One alternative approach is to fix $x + y = s$ and treat $xy$ as $p$. Then the inequality $p > s$ can be rewritten as $p - s > 0$, and using $p \le \frac{s^2}{4}$ by AM-GM, we obtain $s^2/4 - s > 0 \implies s(s - 4) > 0$, which directly gives $s > 4$ since $s > 0$. This approach is algebraically compact but requires recalling the maximum product of two positive numbers with fixed sum. The main approach is preferable because it uses only elementary inequalities without introducing additional parameters and gives a clear step-by-step reasoning, suitable for a contest-style solution.