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 $$(-1{)}^{\lfloor 2^{i+1}a\rfloor }=1-2x_{i+1}.$$ --- 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 $$\begin{array}{rl}36a^2-24a+5+\frac{1}{4^{2019}}& =4(3a-1{)}^2+1+\frac{1}{4^{2019}}\\ & \ge 4(3A+3B-1{)}^2+1+\frac{1}{4^{2019}}.\end{array}$$ We prove that $$\begin{array}{rl}& {\left(2-\frac{1}{2\cdot 4^{1009}}-6A\right)}^2+{\left(1-\frac{1}{4^{1009}}-6B\right)}^2\\ & \le 4(3A+3B-1{)}^2+1+\frac{1}{4^{2019}}.\end{array}$$ 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, $$\begin{array}{rl}& \frac{6}{4^{1009}}A+12B\left(1+\frac{1}{4^{1009}}-6A\right)\\ & \le \frac{6}{4^{1009}}A\le \frac{6}{4^{1009}}\sum _{i=0}^{1009}\frac{1}{2^{2i+1}}=\frac{1}{4^{1008}}-\frac{1}{4^{2018}}.\end{array}$$ 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 $$a=\frac{2}{3}\left(1-\frac{1}{4^{1010}}\right).$$