Find the value of minimize $a+b$ If $a,b$ are positive integers and $$\frac{1}{2a}+\frac{1}{3a}+\frac{1}{4a}=\frac{1}{b^2-2b}$$. Find the value of minimize $a+b$

From $\frac{1}{2a}+\frac{1}{3a}+\frac{1}{4a}=\frac{1}{b^2-2b}\Leftrightarrow13b^2-26b-12a=0$
$\Leftrightarrow 12(a+b)=13b^2-14b$
$\Leftrightarrow a+b=\frac{13b^2-14b}{12}$ 
$\Leftrightarrow a+b=b^2-b+\frac{b^2-2b}{12}=b^2-b+\frac{b\left(b-2\right)}{12}$
We have: $b$ must even number $(1)$
For $\frac{b\left(b-2\right)}{2.2.3}\in Z\Rightarrow b⋮3$ or $b-2⋮3$ $(2)$
From $(1);(2)$ $\Rightarrow b=6k$ or $b-2=6k$ $(k\ge1)$
*)With $b=6k\Rightarrow a+b=\frac{13\left(6k\right)^2-14\cdot 6k}{12}=3k^2-7k$
We have function $f(k)=39k^2-7k$ is function covariate withg $k\ge1$
So $a+b$ have minimize when k have minimize or $k=1$ $\Rightarrow b=6;a=26$
*)With $b-2=6k$.....
this method is inconvenient with me, so i can a new method

Hint:
$12a/13=b(b-2)$
$\implies13|a, a=13c$(say)
$(b-2)b=12c\implies$
$(i) b(b-2)\ge12\implies b>4$
$(ii)b$ must be even $b=2d, d>2$ 
$d(d-1)=3c$
Now $(d,d-1)=1$
