I would like a nudge to the correct answer

$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12}\\ \dfrac{x+y}{xy} = \dfrac{1}{12}\\ \text{Let }x + y = k, xy = 12k.\\ y(k-y) = 12k\\ -y^2 +ky = 12k\\ y^2 -ky+12k = 0\\ y = \dfrac{k+\sqrt{k^2 -48k}}{2}\\ \implies k^2 - 48k \text{ is a perfect square.}\\ k^2 - 48k \ge 0\\ k^2 - 48k = t^2\\ k \le 0\text{(rejected)} \vee k\ge48\\ k \ge 48\\ k \text{ cannot be 48. If } k = 48\text{, } x = y = 24.\\ k \ge 49\\ k = 49 \text{ satisfies the condition that } k^2 - 48k \text{ is a perfect square.}\\ \min(k) = 49\\ \boxed{\min(x + y) = \color{red}49}$

a=1; b=1;c= 1/a + 1/b; if(c==1/12, goto4, goto5);printa, b; a++;if(a<200, goto2, 0);a=1;b++;if(b<200, goto2, discard=0;

a         b

156    13
84      14
60      15
48      16
36      18
30      20
28      21
24      24
21      28
20      30
18      36
16      48
15      60
14      84
13      156

Exactly!.

OHHHH wow... I just checked here and BOOOOMMMM THX GUYS!