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Problem 934

AMC 12 late, AIME early

Algebra Difficulty 4.7 Prove it Austrian Mathematical Olympiad · Austria

Let $a$ , $b$ and $c$ be positive real numbers satisfying $a + b + c + 2 = abc$ .
Prove
$(a + 1)(b + 1)(c + 1) \geq 27.$
When does equality occur?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Answer. Equality occurs if and only if $a = b = c = 2$ .

We set $x = a + 1$ , $y = b + 1$ and $z = c + 1$ . Thus we have to show
$xyz \geq 27$
subject to
$xyz = xy + yz + zx.$
From the constraint we get
$xyz = xy + yz + zx \geq 3\sqrt[3]{x^2y^2z^2}$
by using the inequality between the arithmetic and the geometric means of $xy$ , $yz$ and $zx$ . This is clearly equivalent to $xyz \geq 27$ .
Equality occurs if and only if $xy = yz = zx$ , or, equivalently, $x = y = z$ . By the constraint, this is equivalent to $x = y = z = 3$ and finally $a = b = c = 2$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.