Ta có:
\(\frac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2=2\left(a+b+c\right)\)
=> \(\left(a+b+c\right)^2-6\left(a+b+c\right)\le0\)
=> \(0\le a+b+c\le6.\)
\(T=\frac{a}{a+1}+\frac{b}{b+a}+\frac{c}{c+1}=1-\frac{1}{a+1}+1-\frac{1}{b+1}+1-\frac{1}{c+1}\)
\(=3-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\le3-\frac{\left(1+1+1\right)^2}{a+b+c+3}\le3-\frac{3^2}{6+3}=2\)
"=" xảy ra <=> \(a=b=c\)và \(a+b+c=6\)<=> \(a=b=c=2\)
Vậy max T = 2 khi và chỉ khi a=b=c =2
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