$\frac{a\sqrt{a}}{a+\sqrt{ab}+b}=\frac{\sqrt{a}(a+\sqrt{ab}+b)-\sqrt{ab}(a+b)}{a+\sqrt{ab}+b}=\sqrt{a}-\frac{\sqrt{ab}(a+b)}{a+\sqrt{ab}+b}\geq \sqrt{a}-\frac{\sqrt{ab}(a+b)}{3\sqrt{ab}}=\sqrt{a}-\frac{a+b}{3}$$\Rightarrow VT \geq \sqrt{a}+\sqrt{b}+\sqrt{c}+\frac{1}{27\sqrt{abc}}-\frac{2(a+b+c)}{3}$
$=\frac{1}{3}\sqrt{a}+\frac{1}{3}\sqrt{b}+\frac{1}{3}\sqrt{c}+\frac{1}{27\sqrt{abc}}+\frac{2}{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})-\frac{2}{3} \geq 4\sqrt[4]{\frac{1}{27^2}}+\frac{2}{3}-\frac{2}{3}=\frac{4\sqrt{3}}{9}$
Vì $(a,b,c)\leq 1 \Rightarrow \sqrt{a}\geq a ,\sqrt{b}\geq b, \sqrt{c}\geq c \Rightarrow \sqrt{a}+\sqrt{b}+\sqrt{c}\geq a+b+c=1$