\(T=\frac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\ge\frac{2\left(a+b\right)}{\frac{4a+3a+b}{2}+\frac{4b+3b+a}{2}}=\frac{4\left(a+b\right)}{8\left(a+b\right)}=\frac{1}{2}\)
\(\Rightarrow T_{min}=\frac{1}{2}\) khi \(a=b\)