Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME Prove it Mongolia

For all positive real numbers $u$ and $v$ , prove
$\min \left\{ u, \frac{100}{v}, v + \frac{2023}{u} \right\} \le \sqrt{2123}.$
Here $\min X$ denotes the minimum element of $X$ .

Solution

Let us denote the left side of the given inequality by $S$ .

If $u \le \sqrt{2123}$ , then $S \le u \le \sqrt{2123}$ .

If $\frac{100}{v} \le \sqrt{2123}$ , then $S \le \frac{100}{v} \le \sqrt{2123}$ .

If $u \ge \sqrt{2123}$ and $\frac{100}{v} \ge \sqrt{2123}$ , then we have
$S \le v + \frac{2023}{u} \le \frac{100}{\sqrt{2123}} + \frac{2023}{\sqrt{2123}} = \sqrt{2123}.$
The equality holds when $u = \sqrt{2123}$ and $v = \frac{100}{\sqrt{2123}}$ .

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