$$\dfrac{a}{\sin{A}} = \dfrac{b}{\sin{B}} = \dfrac{c}{\sin{C}} = 2R,$$

trong đó $R$ là bán kính đường tròn ngoại tiếp tam giác $ABC$.

\begin{align*}\sqrt{a\sin{A}} +\sqrt{b\sin{B}} + \sqrt{c\sin{C}} &=\sqrt{2Ra \cdot \frac{a}{2R}} +\sqrt{2Rb \cdot \frac{b}{2R}} +\sqrt{2Rc \cdot \frac{c}{2R}} \\&= \frac{a}{\sqrt{R}} + \frac{b}{\sqrt{R}} + \frac{c}{\sqrt{R}} \\&= \frac{a+b+c}{\sqrt{R}}.\end{align*}

Mặt khác, ta cũng có:

\begin{align*}\sqrt{(a+b+c)(\sin{A} + \sin{B} + \sin{C})} &=\sqrt{(a+b+c) \left(\frac{a}{2R} +\frac{b}{2R} + \frac{c}{2R}\right)} \\&= \sqrt{\frac{(a+b+c)^2}{2R}} \\&= \frac{a+b+c}{\sqrt{R}}.\end{align*}

Do đó, ta có:

$$\dfrac{\sqrt{a\sin{A}} +\sqrt{b\sin{B}} + \sqrt{c\sin{C}}}{\sqrt{(a+b+c)(\sin{A} + \sin{B} +\sin{C})}} = \dfrac{\frac{a+b+c}{\sqrt{R}}}{\frac{a+b+c}{\sqrt{R}}} = \boxed{1}.$$