\(VT\ge\left(a+b+c\right)+\dfrac{9}{a+b+c}\)
\(=\left(a+b+c\right)+\dfrac{1}{9\left(a+b+c\right)}+\dfrac{80}{9\left(a+b+c\right)}\)
\(\ge2\sqrt{\left(a+b+c\right)\dfrac{1}{9\left(a+b+c\right)}}+\dfrac{80}{\dfrac{9.1}{3}}\)\(=\dfrac{82}{3}\)
Vậy ta có đpcm. Đẳng thức xảy ra\(\Leftrightarrow a=b=c=\dfrac{1}{9}\)