Ta có: \(\left(a^2+1\right)\left(\frac{1}{3}+1\right)\ge\left(\frac{a}{\sqrt{3}}+1\right)^2\)
\(\Rightarrow a^2+1\ge\frac{3}{4}\left(\frac{a}{\sqrt{3}}+1\right)^2\Rightarrow\sqrt{a^2+1}\ge\frac{\sqrt{3}}{2}\left(\frac{a}{\sqrt{3}}+1\right)=\frac{1}{2}\left(a+\sqrt{3}\right)\)
\(\Rightarrow M\le\sum\frac{2a}{a+\sqrt{3}}=\sum\frac{2a}{a+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}}\)
\(\Rightarrow M\le\frac{1}{8}\sum a\left(\frac{1}{a}+3\sqrt{3}\right)=\frac{3}{8}+\frac{3\sqrt{3}}{8}\left(a+b+c\right)\le\frac{3}{8}+\frac{3\sqrt{3}}{8}.\sqrt{3}=\frac{3}{2}\)
\(\Rightarrow M_{max}=\frac{3}{2}\) khi \(a=b=c=\frac{1}{\sqrt{3}}\)
