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CM: \(a=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}-\frac{\sqrt{2}}{8}\Rightarrow a+\frac{\sqrt{2}}{8}=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\)
\(\Leftrightarrow\left(a+\frac{\sqrt{2}}{8}\right)^2=\left(\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\right)^2\)\(\Leftrightarrow a^2+\frac{a\sqrt{2}}{4}+\frac{1}{32}=\frac{1}{4}\left(\sqrt{2}+\frac{1}{8}\right)\Leftrightarrow a^2+\frac{2\sqrt{a}}{4}+\frac{1}{32}=\frac{\sqrt{2}}{4}+\frac{1}{32}\)
\(\Leftrightarrow4a^2+\sqrt{2}a-\sqrt{2}=0\)
Theo trên: \(4a^2+\sqrt{2}a-\sqrt{2}=0\Rightarrow a^2=\frac{\sqrt{2}\left(1-a\right)}{4}\Rightarrow a^4=\frac{a^2-2a+1}{8}\)
\(\Rightarrow a^4+a+1=\frac{a^2-2a+1}{8}+a+1=\left(\frac{a+3}{2\sqrt{2}}\right)^2\)
\(B=a^2+\sqrt{a^4+a+1}=a^2+\frac{a+3}{2\sqrt{2}}=\frac{2\sqrt{2}a^2+a+3}{2\sqrt{2}}\)\(=\frac{4a^2+\sqrt{2}a+3\sqrt{2}}{4}=\frac{4\sqrt{2}}{4}=\sqrt{2}\)
3.
\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)
\(=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)
\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)