A problem by Fahim Shahriar Shakkhor

$\angle BAC = 60^\circ$

From Sine Rule,

$\frac{AB}{\sin ACB} = \frac{AC}{\sin ABC} = \frac{BC}{\sin BAC} = 2R$

$\frac{AB}{2R} = \sin ACB$ _ _ _ _ _(1)

$\frac{AC}{2R} = \sin ABC$ _ _ _ _ _(2)

$\frac{BC}{2R} = \sin BAC$ _ _ _ _ _(3)

Add (1),(2),(3).

$\frac{AB + AC + BC}{2R} = \sin ABC + \sin ACB + \sin BAC$

$\frac{AB + AC + BC}{2R} = \sin 45^\circ + \sin (45+30)^\circ + \sin 60^\circ$

$\frac{AB + AC + BC}{2R} = \frac{1}{\sqrt{2}} + \frac{\sqrt{3}+1}{2\sqrt{2}} + \frac{\sqrt{3}}{2}$

$\frac{AB + AC + BC}{2\sqrt{\frac{2}{3}}} = \frac{\sqrt{3}(\sqrt{3} + \sqrt{2} + 1 )}{2\sqrt{2}}$

$AB+BC+CA = \sqrt{3} + \sqrt{2} + 1$

Hence, $a+b+c = 3 + 2 +1 =6$

The circumradius of $\triangle ABC$ is $\sqrt{\dfrac{2}{3}}$ . The angles are $\angle A =60^{\circ}$ , $\angle B = 45^{\circ}$ and $\angle C = 75^{\circ}$ . By the Law of Sines , we have,

$\dfrac{BC}{\sin 60^\circ} = \dfrac{CA}{\sin 45^\circ}=\dfrac{AB}{\sin 75^\circ}=2 \sqrt{\dfrac{2}{3}}$

$\begin{aligned} \therefore ~ BC + CA + AB &=& 2 \sqrt{\dfrac{2}{3}}\left(\sin 60^\circ + \sin 45^\circ + \sin 75^\circ\right) \\ \\ &=& 2 \sqrt{\dfrac{2}{3}}\left(\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt 2 + \sqrt 6}{4}\right) \\ \\ &=& \dfrac{\sqrt2}{\sqrt3}\times\dfrac{2\sqrt{3} + 3\sqrt 2 + \sqrt 6 }{2} \\ \\ &=& \dfrac{2\sqrt{3} + 3\sqrt 2 + \sqrt 6 }{\sqrt{6}} \\ \\ &=& \dfrac{2\sqrt{3}}{\sqrt{6}} + \dfrac{3\sqrt{2}}{\sqrt{6}} + \dfrac{\sqrt{6}}{\sqrt{6}} \\ \\ &=& \sqrt2 + \sqrt3 + 1 \\ \\ \end{aligned}$

Implying $a=2$ , $b=3$ , $c=1$ , giving our desired answer:

$a+b+c=2+3+1=\fbox{6}$