Iveela wrote: Let $u, v$ be arbitrary positive real numbers. Prove that \[\min{(u, \frac{100}{v}, v+\frac{2023}{u})} \leq \sqrt{2123}.\] If $u\le\sqrt{2123}$ : $\min(u,\frac {100}v,v+\frac{2023}u)\le u\le\sqrt{2123}$ If $v\ge\frac{100}{\sqrt{2123}}$ : $\min(u,\frac {100}v,v+\frac{2023}u)\le \frac {100}v\le\sqrt{2123}$ If $u>\sqrt{2123}$ and $v<\frac{100}{\sqrt{2123}}$ : $\min(u,\frac {100}v,v+\frac{2023}u)\le v+\frac{2023}u<\frac{100}{\sqrt{2123}}+\frac{2023}{\sqrt{2123}}=\sqrt{2123}$

Iveela wrote: Let $u, v$ be arbitrary positive real numbers. Prove that \[\min{(u, \frac{100}{v}, v+\frac{2023}{u})} \leq \sqrt{2123}.\] Interesting.

Attachments:

Iveela wrote: Let $u, v$ be arbitrary positive real numbers. Prove that \[\min{(u, \frac{100}{v}, v+\frac{2023}{u})} \leq \sqrt{2123}.\] Solution.if $u>\sqrt{2123}$ and $\frac{100}{v}>\sqrt{2123}$, then \begin{align*}& v+ \frac{2023}{u}<\frac{100}{\sqrt{2123}}+ \frac{2023}{\sqrt{2123}} =\sqrt{2123}; \end{align*}otherwise, one of $u$ and $\frac{100}{v}\le\sqrt{2123}$, hence the result. $\blacksquare$