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Problem 957

AIME late

Algebra Difficulty 5.9 Prove it

Example 1 Given real numbers $x, y$ satisfy $x+y=1$ , prove:
$x y \leqslant \frac{1}{4}$

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

Proof: Since $x+y=1$ , by the pigeonhole principle, it is known that $x, y$ are on opposite sides of $\frac{1}{2}$ (or at $\frac{1}{2}$ ),

i.e., $\left(x-\frac{1}{2}\right)\left(y-\frac{1}{2}\right) \leqslant 0$ , thus $x y-\frac{1}{2}(x+$ $y)+\frac{1}{4} \leqslant 0 \Leftrightarrow x y \leqslant \frac{1}{2}(x+y)-\frac{1}{4} \Leftrightarrow x y \leqslant \frac{1}{4}$ .

Therefore, the original inequality holds, with equality if and only if $x=y=\frac{1}{2}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.