\(a=\frac{1}{8}\left(4\sqrt{\sqrt{2}+\frac{1}{8}}-\sqrt{2}\right)>\frac{1}{8}\left(4-\sqrt{2}\right)>0\)
\(a^2=\frac{1}{4}\left(\sqrt{2}+\frac{1}{8}\right)+\frac{1}{32}-\frac{\sqrt{2}}{8}\sqrt{\sqrt{2}+\frac{1}{8}}\)
\(=\frac{\sqrt{2}}{4}+\frac{1}{16}-\frac{\sqrt{2}}{8}\sqrt{\sqrt{2}+\frac{1}{8}}\)
\(=\frac{\sqrt{2}}{4}-\frac{\sqrt{2}}{4}\left(\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}-\frac{1}{8}\sqrt{2}\right)\)
\(=\frac{\sqrt{2}}{4}-\frac{\sqrt{2}}{4}a< \frac{\sqrt{2}}{4}< \sqrt{2}\)
\(\Rightarrow2\sqrt{2}a^2=1-a\Rightarrow a=1-2\sqrt{2}a^2\)
\(\Rightarrow A=a^2+\sqrt{a^4+a+1}=a^2+\sqrt{a^4-2\sqrt{2}a^2+2}\)
\(=a^2+\sqrt{\left(a^2-\sqrt{2}\right)^2}=a^2+\left|a^2-\sqrt{2}\right|=a^2+\sqrt{2}-a^2=\sqrt{2}\)
