Ta có:
\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2017\)
\(\Leftrightarrow7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=20\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2017\le\frac{20}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2017\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{6051}\)
Ta lại có:
\(T=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
\(\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}+\frac{2}{b}+\frac{1}{c}+\frac{2}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{\sqrt{6051}}{3}\)
