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Problem 921

AIME late

Algebra Difficulty 5.8 Prove it

One, (40 points) Let complex numbers $u, v$ satisfy
$\begin{array}{l} 3|u+1||v+1| \geqslant|u v+5 u+5 v+1|, \\ |u+v|=|u v+1| . \end{array}$

Prove: One of $u, v$ must be equal to 1.

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

Let $u+v$ and $uv+1$ be denoted as $s$ and $t$ respectively.
Then, from the conditions, we have $|s|=|t|$ , and
$\begin{array}{l} |5 s+t| \leqslant 3|s+t| . \\ \text { Let }|s|^{2}=|t|^{2}=A, \frac{1}{2}(s \bar{t}+\bar{s} t)=B \text {. Then } \\ 26 A+10 B=|5 s+t|^{2} \\ \leqslant 9|s+t|^{2}=9(2 A+2 B) \\ \Rightarrow A \leqslant B . \end{array}$

At this point, $|s-t|^{2}=2(A-B) \leqslant 0$ , which can only be $s=t$ . Therefore, $(u-1)(v-1)=t-s=0$ , meaning that one of $u$ or $v$ must be equal to 1.

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