\(G=\dfrac{1}{\sqrt{6}}.\sqrt{6}.\sqrt{15a+1}+\dfrac{1}{\sqrt{6}}.\sqrt{6}.\sqrt{15b+1}+\dfrac{1}{\sqrt{6}}.\sqrt{6}.\sqrt{15c+1}\)
\(G\le\dfrac{1}{2\sqrt{6}}\left(6+15a+1+6+15b+1+6+15c+1\right)=3\sqrt{6}\)
\(G_{max}=3\sqrt{6}\) khi \(a=b=c=\dfrac{1}{3}\)
Do \(\left\{{}\begin{matrix}a;b;c\ge0\\a+b+c=1\end{matrix}\right.\) \(\Rightarrow0\le a;b;c\le1\)
\(\Rightarrow\left\{{}\begin{matrix}a\left(1-a\right)\ge0\\b\left(1-b\right)\ge0\\c\left(1-c\right)\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a\ge a^2\\b\ge b^2\\c\ge c^2\end{matrix}\right.\)
\(\Rightarrow\sum\sqrt{15a+1}=\sum\sqrt{9a+6a+1}\ge\sum\sqrt{9a^2+6a+1}=3\left(a+b+c\right)+3=6\)
\(G_{min}=6\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị
