22nd Chinese Girls' Mathematical Olympiad

Let $y_i=\frac{x_i}{2^{i/2}}$, for $i=1,2,\dots ,22$. It is well-known that $f(t)=t+\frac{1}{t}$ is decreasing on $(0,1]$ and increasing on $[1,+\infty )$. For $1\le i\le 11$, we have $\frac{1}{2^{12-i}}\le y_i\le \frac{1}{2^{11-i}}$, thus $y_i+\frac{1}{y_i}\le 2^{12-i}+\frac{1}{2^{12-i}}$. For $12\le i\le 22$, we have $2^{i-12}\le y_i\le 2^{i-11}$, thus $y_i+\frac{1}{y_i}\le 2^{i-11}+\frac{1}{2^{i-11}}$. Hence, we have $$\begin{array}{rl}\left(\sum _{i=1}^{22}{x}_i\right)\left(\sum _{i=1}^{22}\frac{1}{x_i}\right)& =\left(\sum _{i=1}^{22}{y}_i\right)\left(\sum _{i=1}^{22}\frac{1}{y_i}\right)\\ & \le \frac{1}{4}{\left(\sum _{i=1}^n\left(y_i+\frac{1}{y_i}\right)\right)}^2\\ & \le \frac{1}{4}{\left(\sum _{i=1}^{11}\left(2^{12-i}+\frac{1}{2^{12-i}}\right)+\sum _{i=12}^{22}\left(2^{i-11}+\frac{1}{2^{i-11}}\right)\right)}^2\\ & ={\left(2^1+2^2+\cdots +2^{11}+\frac{1}{2^1}+\frac{1}{2^2}+\cdots +\frac{1}{2^{11}}\right)}^2\\ & ={\left(2^{12}-1-\frac{1}{2^{11}}\right)}^2.\end{array}$$ When $$x_i=\{\begin{array}{ll}{2}^{i-1},& 1\le i\le 11,\\ 2^i,& 12\le i\le 22,\end{array}$$ equality holds. Thus, the maximum value sought is $(2^{12}-1-\frac{1}{2^{11}}{)}^2$. $\square$