VN IMO Booklet

(a) By assumption, we have $v_i = u_i$ for even $i$ and $v_i = -u_i$ for odd $i$. Thus, the inequality can be rewritten as
$$\left(\sum_{i=0}^{1009} u_{2i} + \sum_{i=0}^{1008} u_{2i+1}\right)^2 + \left(\sum_{i=0}^{1009} u_{2i} - \sum_{i=0}^{1008} u_{2i+1}\right)^2 \le 72a^2 - 48a + 10 + \frac{2}{4^{2019}},$$
which is equivalent to
$$\left(\sum_{i=0}^{1009} u_{2i}\right)^2 + \left(\sum_{i=0}^{1008} u_{2i+1}\right)^2 \le 36a^2 - 24a + 5 + \frac{1}{4^{2019}}.$$
Let the binary representation of $a$ be $a = \sum_{i=1}^{+\infty} \frac{x_i}{2^i}$ where $x_i \in \{0, 1\}$. Since $\frac{1}{2} \le a \le \frac{2}{3}$, we have $x_1 = 1$.
For each natural number $i$, the parity of $\lfloor 2^{i+1}a \rfloor$ depends on $x_{i+1}$. In particular, if $x_{i+1} = 0$, then $\lfloor 2^{i+1}a \rfloor$ is even, and if $x_{i+1} = 1$, then $\lfloor 2^{i+1}a \rfloor$ is odd. Therefore, $(-1)^{\lfloor 2^{i+1}a \rfloor} = 1$ if $x_{i+1} = 0$, and $(-1)^{\lfloor 2^{i+1}a \rfloor} = -1$ if $x_{i+1} = 1$. In all cases, we have
$$(-1)^{\lfloor 2^{i+1}a \rfloor} = 1 - 2x_{i+1}.$$
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Denote $A = \sum_{i=0}^{1009} \frac{x_{2i+1}}{2^{2i+1}}$ and $B = \sum_{i=0}^{1008} \frac{x_{2i+2}}{2^{2i+2}}$, we have
$$\sum_{i=0}^{1009} u_{2i} = \sum_{i=0}^{1009} \frac{3(1 - 2x_{2i+1})}{2^{2i+1}} = 2 - \frac{1}{2 \cdot 4^{1009}} - 6A,$$
$$\sum_{i=0}^{1008} u_{2i+1} = \sum_{i=0}^{1008} \frac{3(1 - 2x_{2i+2})}{2^{2i+2}} = 1 - \frac{1}{4^{1009}} - 6B.$$
On the other hand, $a \ge A + B \ge \frac{1}{2}$, thus
$$\begin{aligned} 36a^2 - 24a + 5 + \frac{1}{4^{2019}} &= 4(3a - 1)^2 + 1 + \frac{1}{4^{2019}} \\ &\ge 4(3A + 3B - 1)^2 + 1 + \frac{1}{4^{2019}}. \end{aligned}$$
We prove that
$$\begin{aligned} & \left(2 - \frac{1}{2 \cdot 4^{1009}} - 6A\right)^2 + \left(1 - \frac{1}{4^{1009}} - 6B\right)^2 \\ & \le 4(3A + 3B - 1)^2 + 1 + \frac{1}{4^{2019}}. \end{aligned}$$
By some calculations, we can rewrite the above inequality as
$$\frac{6}{4^{1009}}A + 12B \left(1 + \frac{1}{4^{1009}} - 6A\right) \le \frac{1}{4^{1008}} - \frac{1}{4^{2018}}.$$
Since $A \ge \frac{1}{2}$, we have $6A > 1 + \frac{1}{4^{1009}}$. Thus,
$$\begin{aligned} & \frac{6}{4^{1009}}A + 12B \left(1 + \frac{1}{4^{1009}} - 6A\right) \\ & \le \frac{6}{4^{1009}}A \le \frac{6}{4^{1009}} \sum_{i=0}^{1009} \frac{1}{2^{2i+1}} = \frac{1}{4^{1008}} - \frac{1}{4^{2018}}. \end{aligned}$$
Using these inequalities, the desired result will follow.

(b) By the arguments in part (a), the equality holds if and only if $a = A + B, B = 0$ and $A = \frac{2}{3}\left(1 - \frac{1}{4^{1010}}\right)$, which implies that
$$a = \frac{2}{3} \left( 1 - \frac{1}{4^{1010}} \right).$$