I haven't work in detail yet but strongly believe the following facts are the only main ideas of the problem: Let the binary representation of $a$ be $\sum_{i=0}^{\infty}{\frac{d_i}{2^i}}$ where $d_i\in \{ 0,1\}$, it's enough to consider only when $a= \sum_{i=0}^{2019}{\frac{d_i}{2^i}}$. For each $i\in \{ 0,1,2,\dotsc ,2018\}$, $$(-1)^{\lfloor 2^{i+1}a\rfloor} =\begin{cases} 1, &\text{ if }d_{i+1}=0 \\ -1, &\text{ if }d_{i+1}=1 \end{cases}\implies (-1)^{\lfloor 2^{i+1}a\rfloor} =-2d_{i+1}+1.$$

Here is my solution for this problem Solution a) We have: $u_i = v_i$ if $i$ is even, $u_i = - v_i$ if $i$ is odd Then the inequality becomes: $\left[\sum^{1009}_{i = 0} (u_{2i} + u_{2i + 1})\right]^2 + \left[\sum^{1008}_{i = 0} (u_{2i} - u_{2i + 1})\right]^2 \le 72a^2 - 48a + 10 + \dfrac{2}{4^{2019}}$ $\Leftrightarrow$ $\left(\sum^{1009}_{i = 0} u_{2i}\right)^2 + \left(\sum^{1008}_{i = 0} u_{2u + 1}\right)^2 \le 36a^2 - 24a + 5 + \dfrac{1}{4^{2019}}$ Let $a = \sum^{\infty}_{i = 0} \dfrac{x_i}{2^i}$ with $x_i$ $\in$ $\{0; 1\}$ As ThE-dArK-lOrD, $(- 1)^{[2^{i + 1}a]} = 1 - 2x_{i + 1}$ Let $P = \sum^{1009}_{i = 0} \dfrac{x_{2i + 1}}{2^{2i + 1}}$, $Q = \sum^{1008}_{i = 0} \dfrac{x_{2i + 2}}{2^{2i + 2}}$ We have: $u_{2i} = \dfrac{3 - 6x_{2i + 1}}{2^{2i + 1}}, u_{2i + 1} = \dfrac{3 - 6x_{2i + 2}}{2^{2i + 2}}$ So: $\sum^{1009}_{i = 0} u_{2i} = 2 - \dfrac{1}{2 . 4^{1009}} - 6P$, $\sum^{1008}_{i = 0} u_{2i + 1} = 1 - \dfrac{1}{4^{1009}} - 6Q$ Combine with: $\dfrac{2}{3} \ge a \ge P + Q \ge \dfrac{1}{2}$, the inequality becomes: $36(P + Q)^2 - 24(P + Q) + 5 + \dfrac{1}{4^{2019}} \ge \left(2 - \dfrac{1}{2 . 4^{1009}} - 6P\right)^2 + \left(1 - \dfrac{1}{4^{1009}} - 6Q\right)^2$ $\Leftrightarrow$ $\dfrac{1}{4^{1008}} - \dfrac{1}{4^{2018}} \ge \dfrac{6P}{4^{1009}} + 12Q\left(1 + \dfrac{1}{4^{1009}} - 6P\right)$ $(1)$ We have: $\dfrac{6}{4^{1009}}P + 12Q\left(1 + \dfrac{1}{4^{1009}} - 6P\right) \le \dfrac{6}{4^{1009}}P \le \dfrac{6}{4^{1009}} \left(\sum^{1009}_{i = 0} \dfrac{1}{2^{2i + 1}}\right) = \dfrac{6}{4^{1009}} \left(\dfrac{2}{3} - \dfrac{2}{3 . 4^{1010}}\right) = \dfrac{1}{4^{1008}} - \dfrac{1}{4^{2018}}$ It's mean $(1)$ is true Hence the inequality has been proved b) The equality holds when: $a = P + Q$, $Q = 0$ or $a = P = \dfrac{2}{3} - \dfrac{2}{3 . 4^{1010}}$

khanhnx wrote: $\dfrac{6}{4^{1009}}P + 12Q\left(1 + \dfrac{1}{4^{1009}} - 6P\right) \le \dfrac{6}{4^{1009}}P$ why?

aops_user_0 wrote: khanhnx wrote: $\dfrac{6}{4^{1009}}P + 12Q\left(1 + \dfrac{1}{4^{1009}} - 6P\right) \le \dfrac{6}{4^{1009}}P$ why? Because $P \ge \dfrac{1}{2}$