Complex Fractional Equation

Set $a=y+z-2, b=z+x-4$ and $c=x+y-6$ . Then we have that $a,b,c>0$ since $x,y,z>3$ and that $x+2 = \frac{b+c-a+12}{2}, y+4=\frac{c+a-b+12}{2}$ and $z+6=\frac{a+b-c+12}{2}$ . Thus, rewriting the given equation and using Cauchy-Schwarz Inequality we get $\begin{aligned} 36&=\dfrac{(b+c-a+12)^2}{4a}+\dfrac{(c+a-b+12)^2}{4b}+\dfrac{(a+b-c+12)^2}{4c}\\ &\geq \dfrac{(a+b+c+36)^2}{4(a+b+c)}, \end{aligned}$ which follows that $(a+b+c+36)^2\leq 144(a+b+c)\iff (a+b+c-36)^2 \leq 0.$ Since $(a+b+c-36)^2 \geq 0$ , we must have $(a+b+c-36)^2 = 0$ , so that $a+b+c=36$ . Furthermore, equality holds when $\dfrac{b+c-a+12}{4a}=\dfrac{c+a-b+12}{4b}=\dfrac{a+b-c+12}{4c},$ which simplifies to $a=b=c$ . Thus, we have $a=b=c=12$ , so $y+z=14, z+x=16$ and $x+y=18$ . Hence, we get $x=\frac{16+18-14}{2}=\boxed{10}$ .

It is given that there is only one triplet $(x,y,z)$ that satisfies the equation. Looking at the equation and to simply it, we assume that $y=x-2$ and $z=x-4$ , then:

$\quad \dfrac {(x+2)^2}{y+z-2} + \dfrac {(y+4)^2}{z+x-4} + \dfrac {(z+6)^2}{x+y-6} = 36$

$\quad \Rightarrow \dfrac {(x+2)^2}{x-2+x-4-2} + \dfrac {(x-2+4)^2}{x-4+x-4} + \dfrac {(x-4+6)^2}{x+x-2-6} = 36$

$\quad \Rightarrow \dfrac {(x+2)^2}{2x-8} + \dfrac {(x+2)^2}{2x-8} + \dfrac {(x+2)^2}{2x-8} = 36$

$\quad \Rightarrow \dfrac {(x+2)^2}{2x-8} = 12 \quad \Rightarrow x^2+4x+4 = 24x-96$

$\quad \Rightarrow x^2-20x+100 = 0 \quad \Rightarrow (x-10)^2 = 0 \quad \Rightarrow x = \boxed{10}$

i did something which i am myself not able to understand but i got answer as 10 alright.........i sub x+2=u,y+4=v,z+6=w ...............and then i used symmetry with titu's lemma and got x=10 . cheers