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Problem 486

AMC 12 late, AIME early

Algebra Difficulty 4.9 Prove it National Math Olympiad · Slovenia

Prove that for all real $x$ and $y$ the inequality
$|x + y| + |x + 1| + |y + 1| \geq 2$
holds. For what $x$ does there exist $y$ such that $|x+y|+|x+1|+|y+1| = 2$ ?

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

For all real $a$ we have $|a| \ge a$ and $|a| \ge -a$ . So,
$|x + y| \ge -(x + y), \quad |x + 1| \ge x + 1, \quad |y + 1| \ge y + 1, \quad (1)$

$|x + y| + |x + 1| + |y + 1| \geq -(x + y) + (x + 1) + (y + 1) = 2.$
$Assume that the equality holds. Then the equality case occurs in all three inequalities (1) and so $x + y \leq 0$, $x + 1 \geq 0$ and $y + 1 \geq 0$. We get $-1 \leq x \leq -y \leq 1$ or $-1 \leq x \leq 1$. If $-1 \leq x \leq 1$ and $y = -x$, then$
|x + y| + |x + 1| + |y + 1| = 2.

We can conclude that only for all real $x \in [-1, 1]$ there exists $y$ , such that the equality holds.

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