Because $a$, $b$, $c$ are sides of a triangle, we can let $ a=x+y$, $ b=y+z$ and $ c=z+x$ where $ x+y+z=\frac{3}{2}$. Then, \begin{align*} & a^{2}+b^{2}+c^{2}+\frac{4abc}{3}\\ =\quad&\frac{3\cdot(a^{2}+b^{2}+c^{2})+4abc}{3}\\ =\quad&\frac{(a+b+c)(a^{2}+b^{2}+c^{2})+4abc}{3}\\ =\quad&\frac{2(x+y+z)((x+y)^{2}+(y+z)^{2}+(z+x)^{2})+4(x+y)(y+z)(z+x)}{3}\\ =\quad&\frac{4(x^{3}+y^{3}+z^{3}+3x^{2}y+3xy^{2}+3y^{2}z+3yz^{2}+3z^{2}x+3zx^{2}+5xyz)}{3}\\ =\quad&\frac{4((x+y+z)^{3}-xyz)}{3}\\ =\quad&\frac{4(\frac{26}{27}(x+y+z)^{3}+(\frac{x+y+z}{3})^{3}-xyz)}{3}\\ \geq\quad&\frac{4(\frac{26}{27}(x+y+z)^{3}}{3}\\ =\quad&\boxed{\frac{13}{3}} \end{align*}