Ta có $a+b+c=3$ nên

$\begin{array}{l} \dfrac{{\sqrt {3a + bc} }}{{a + \sqrt {3a + bc} }} = \dfrac{{\sqrt {\left( {a + b + c} \right)a + bc} }}{{a + \sqrt {\left( {a + b + c} \right)a + bc} }}\\  = \dfrac{{\sqrt {{a^2} + ab + ac + bc} }}{{a + \sqrt {{a^2} + ab + ac + bc} }}\\  = \dfrac{{\sqrt {\left( {a + b} \right)\left( {a + c} \right)} }}{{a + \sqrt {\left( {a + b} \right)\left( {a + c} \right)} }} = \dfrac{1}{{\dfrac{a}{{\sqrt {\left( {a + b} \right)\left( {a + c} \right)} }} + 1}} \end{array}$

Tương tự ta có:

$\begin{array}{l} \dfrac{{\sqrt {3b + ac} }}{{b + \sqrt {3b + ac} }} = \dfrac{1}{{\dfrac{b}{{\sqrt {\left( {b + c} \right)\left( {b + a} \right)} }} + 1}}\\ \dfrac{{\sqrt {3c + ab} }}{{c + \sqrt {3c + ab} }} = \dfrac{1}{{\dfrac{c}{{\sqrt {\left( {c + a} \right)\left( {c + b} \right)} }} + 1}} \end{array}$

Áp dụng bất đẳng thức $B-C-S$ dạng Engel ta được:

$\begin{array}{l} \dfrac{1}{{\dfrac{a}{{\sqrt {\left( {a + b} \right)\left( {a + c} \right)} }} + 1}} + \dfrac{1}{{\dfrac{b}{{\sqrt {\left( {b + c} \right)\left( {b + a} \right)} }} + 1}} + \dfrac{1}{{\dfrac{c}{{\sqrt {\left( {c + a} \right)\left( {c + b} \right)} }} + 1}}\\  \ge \dfrac{9}{{3 + \dfrac{a}{{\sqrt {\left( {a + b} \right)\left( {a + c} \right)} }} + \dfrac{b}{{\sqrt {\left( {b + c} \right)\left( {b + a} \right)} }} + \dfrac{c}{{\sqrt {\left( {c + a} \right)\left( {c + b} \right)} }}}} \end{array}$

Áp dụng bất đẳng thức $AM-GM$ ta được:

$\begin{array}{l}  \dfrac{a}{{\sqrt {\left( {a + b} \right)\left( {a + c} \right)} }} + \dfrac{b}{{\sqrt {\left( {b + c} \right)\left( {b + a} \right)} }} + \dfrac{c}{{\sqrt {\left( {c + a} \right)\left( {c + b} \right)} }}\\  \le \dfrac{1}{2}\left( {\dfrac{a}{{a + b}} + \dfrac{a}{{a + c}} + \dfrac{b}{{b + c}} + \dfrac{b}{{b + a}} + \dfrac{c}{{c + a}} + \dfrac{c}{{c + b}}} \right) = \dfrac{1}{2}\left( {1 + 1 + 1} \right) = \dfrac{3}{2}\\  \Rightarrow 0 < 3 + \dfrac{a}{{\sqrt {\left( {a + b} \right)\left( {a + c} \right)} }} + \dfrac{b}{{\sqrt {\left( {b + c} \right)\left( {b + a} \right)} }} + \dfrac{c}{{\sqrt {\left( {c + a} \right)\left( {c + b} \right)} }} \le 3 + \dfrac{3}{2} = \dfrac{9}{2} \end{array}$

$ \Leftrightarrow \dfrac{9}{{3 + \dfrac{a}{{\sqrt {\left( {a + b} \right)\left( {a + c} \right)} }} + \dfrac{b}{{\sqrt {\left( {b + c} \right)\left( {b + a} \right)} }} + \dfrac{c}{{\sqrt {\left( {c + a} \right)\left( {c + b} \right)} }}}} \ge \dfrac{9}{{\dfrac{9}{2}}} = 2$

Vậy: $\dfrac{{\sqrt {3a + bc} }}{{a + \sqrt {3a + bc} }} + \dfrac{{\sqrt {3b + ac} }}{{b + \sqrt {3b + ac} }} + \dfrac{{\sqrt {3c + ab} }}{{c + \sqrt {3c + ab} }} \ge 2$