First Problem for Floors

By Cauchy Schwartz,

$\large {(\lfloor\sqrt{3x}\rfloor + \lfloor\sqrt{4y}\rfloor)^2 \leq 2 \left( (\lfloor\sqrt{3x}\rfloor)^2 + (\lfloor\sqrt{4y}\rfloor)^2 \right) \leq 2(\lfloor3x\rfloor + \lfloor4y\rfloor) \leq 2(3x+4y) \leq 2(10) \\ \implies \lfloor\sqrt{3x}\rfloor + \lfloor\sqrt{4y}\rfloor \leq \sqrt{20}}$

$\large {\text{From the inequality above, we can say that:} \\ (\lfloor\sqrt[n]{3x}\rfloor + \lfloor\sqrt[n]{4y}\rfloor)^2 \leq 2(\lfloor\sqrt[\frac{n}{2}]{3x}\rfloor + \lfloor\sqrt[\frac{n}{2}]{4y}\rfloor)}$

$\large{ (\lfloor\sqrt[4]{3x}\rfloor + \lfloor\sqrt[4]{4y}\rfloor)^2 \leq 2(\lfloor\sqrt{3x}\rfloor + \lfloor\sqrt{4y}\rfloor) \leq 2(\sqrt{20}) \\ \implies \lfloor\sqrt[4]{3x}\rfloor + \lfloor\sqrt[4]{4y}\rfloor \leq \sqrt{2\sqrt{20}}}$

$\large{ (\lfloor\sqrt[8]{3x}\rfloor + \lfloor\sqrt[8]{4y}\rfloor)^2 \leq 2(\lfloor\sqrt[4]{3x}\rfloor + \lfloor\sqrt[4]{4y}\rfloor) \leq 2(\sqrt{2\sqrt{20}}) \\ \implies \lfloor\sqrt[8]{3x}\rfloor + \lfloor\sqrt[8]{4y}\rfloor \leq \sqrt{2\sqrt{2\sqrt{20}}}} \\ \cdots$

Hence,

$\large {\max(\lfloor\sqrt{3x}\rfloor + \lfloor\sqrt{4y}\rfloor + \lfloor\sqrt[4]{3x}\rfloor + \lfloor\sqrt[4]{4y}\rfloor + \lfloor\sqrt[8]{3x}\rfloor + \lfloor\sqrt[8]{4y}\rfloor + \lfloor\sqrt[16]{3x}\rfloor + \lfloor\sqrt[16]{4y}\rfloor + \cdots + \lfloor\sqrt[512]{3x}\rfloor + \lfloor\sqrt[512]{4y}\rfloor) \\ = \lfloor\sqrt{20}\rfloor + \lfloor \sqrt{2\sqrt{20}} \rfloor + \lfloor \sqrt{2\sqrt{2\sqrt{20}}} \rfloor + ... + \underbrace{\lfloor \sqrt{2\sqrt{2\sqrt{...\sqrt{20}}}} \rfloor }_{8 \ 2's \ \text{(not considering the digit 2 in 20)}} \\ = 4 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 \\ = \ \boxed{20}}$