Its beautiful (Just observe it)

$x = tana , y = tanb , z = tanc$

$xyz = x + y+ z$ keeping the values it shows the property of a triangle

$tanatanbtanc = tana + tanb + tanc$

$- tana = \frac{tanb + tanc}{1- tanbtanc} = tan(b + c)$

$tan(kpi - a) = tan( b + c)$

$a + b + c = kpi$

$4\frac{\sqrt{ tan^{2}a + 1}}{tana} = ......$

$\frac{4}{sina} = \frac{5}{sinb} = \frac{6}{sinc}$

for k = 1 there is a triangle whose sides are 4k , 5k and 6k

$s = \frac{15}{2}$

$tan\frac{a}{2} = \sqrt{\frac{(s - 5k)(s - 6k)}{s(s - 4k)}} = \sqrt{\frac{1}{7}}$

$x = tana = \frac{2t}{1 - t^{2}} = \frac{\sqrt{7}}{3}$ t = tana/2

similarly , $y =\frac{5\sqrt{7}}{9} , z = 3\sqrt{7}$

x/√(x x+1)=sinX , y/√(y y+1)=sinY , z/√(z z+1)=sinZ. Now we have 4/sinX=5/sinY=6/sinZ,and this equations are similiar as sine law for triangle which sides are 4,5,6(4k,5k,6k exacly but we get least value,it's easier to calculte). So X,Y,Z are angles of tringle whisch sides are 4,5,6.From cosine law we found that cosX=(6 6+5 5-4 4)/(2 6 5)=3/4 so sinX=√(1-9/16)=√7/4.Same cosY=(6 6+4 4-5 5)/(2 4 6)=9/16,and sinY=√(1-81/256)=5√7/16,same way we found cosZ=(4 4+5 5-6 6)/(2 4 5)=1/8,and sinZ =3√7/8. now we replace that values and get x/√(x x+1)=√7/4,so 4x=√(7x x+7),when squared 16xx=7xx+7 , 9xx=7 , xx=7/9 , x=√7/3,similiar y/√(y y+1)=5√7/16, so 16y=√(175yy+175),when squared 256yy=175yy+175, which means yy=175/81,y=5√7/9, and same way z/√(z z+1)=3√7/8, meaning 8z=√(63zz+63),squared again 64zz=63zz+63,zz=63,z=3√7. We found that x=√7/3 , y=5√7/9 , z=3√7. x+y+z=√7(1/3+5/9+3)=√7(3+5+27)/9=35√7/9 xyz=(√7/3)(5√7/9) 3√7=5 7√7/9=35√7/9,so xyz=x+y+z and our x,z,y are solutions of this problem.It means a=7,b=3,c=5,d=7,e=9,f=3,g=7,so a+b+c+d+e+f+g=7+3+5+7+9+3+7=41.