\(Q=\sqrt{\left(a+b+c\right)a+bc}+\sqrt{\left(a+b+c\right)b+ca}+\sqrt{\left(a+b+c\right)c+ab}=\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}\le\dfrac{a+b+a+c+b+a+b+c+c+a+c+b}{2}=\dfrac{4\left(a+b+c\right)}{2}=2\left(a+b+c\right)=4\)
Dấu "=" xảy ra <=> \(a=b=c=\dfrac{2}{3}\)