Three Variables, One Equation!

If we let $x-1=a$ , $y-1=b$ , $z-1=c$ and rewrite the equation, we have,

$a+b+c+ \dfrac3{a} + \dfrac3{b} +\dfrac3{c}+3 = 2 \left( \sqrt{a+3} + \sqrt{b+3} + \sqrt{c+3} \right)$

Let $f(t)=t+\frac{3}{t}-2\sqrt{t+3}+1$ , the equation is now,

$f(a)+f(b)+f(c)=0$

Note that $f(t) \geq 0 \ \forall \ t \geq 0$

$\therefore f(a)=f(b)=f(c)=0$

If $f(t)=0$

$\implies t + \dfrac{3}{t} +1 =2\sqrt{t+3}$

$\implies (t^2-t-3)^2=0$

$\implies t=\dfrac{1}{2} (1+\sqrt{13})$

$\implies x+y+z=\dfrac{3}{2} (3+\sqrt{13})$

$\implies A+B+C+D=\boxed{21}$

(x-1) + (y-1) + (z-1) + [3/(x-1) +1] + [3/(y-1) + 1] + [3/(z-1) + 1] = 2 [(x+2)^1/2 + (y+2)^1/2 + (z+2)^1/2]

(x-1) + (y-1) + (z-1) + (x+2)/(x-1) + (y+2)/(y-1) + (z+1)/(z-1) = 2 [(x+2)^1/2 + (y+2)^1/2 + (z+2)^1/2]

applying A.M.-G.M. for 1st and 4th term, 2nd and 5th term, 3rd and 6th term, we get that equality holds in A.M.-G.M., which means,

x-1 = (x+2)/(x-1)

solving the quadratic we get the required answer keeping the condition in mind x,y,z>=1