Không mất tính tổng quát, giả sử \(a\le b\le c\)
Nếu \(a=b=c\Rightarrow T=0\)
Nếu \(a=b=0,c>0\Rightarrow c=2\Rightarrow T=1\)
Nếu \(\left\{{}\begin{matrix}a=0\\b;c>0\end{matrix}\right.\) \(\Rightarrow2\left(b+c\right)=b^2+c^2\ge\frac{\left(b+c\right)^2}{2}\Rightarrow b+c\le4\)
\(T=\frac{b}{\frac{b}{2}+\frac{b}{2}+1}+\frac{c}{\frac{c}{2}+\frac{c}{2}+1}\le\frac{1}{9}\left(2+2+b+2+2+c\right)\le\frac{1}{9}.12=\frac{4}{3}\)
Nếu a;b;c dương:
\(2\left(a+b+c\right)=a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)
\(\Rightarrow a+b+c\le6\)
\(T=\frac{a}{\frac{a}{2}+\frac{a}{2}+1}+\frac{b}{\frac{b}{2}+\frac{b}{2}+1}+\frac{c}{\frac{c}{2}+\frac{c}{2}+1}\le\frac{1}{9}\left(4+a+4+b+4+c\right)\)
\(T\le\frac{1}{9}\left(12+a+b+c\right)\le2\)
So sánh các giá trị ta thấy \(T_{max}=2\) khi \(a=b=c=2\)
