For a>b>c>0, the distance between (1,1) and the point of intersection of the lines ax+by+c=0 and bx+ay+c=0 is less than 2√2, then

The point of intersection for ax+by+c=0 and bx+ay+c=0 is (-c/(a+b), -c/(a+b)).
The distance between (1,1) and (-c/(a+b), -c/(a+b)) is given by:
√[(1 + c/(a+b))^2 + (1 + c/(a+b))^2] = √[2(1 + c/(a+b))^2] = √2 |1 + c/(a+b)|
Since the distance is less than 2√2, we have:
√2 |1 + c/(a+b)| < 2√2
|1 + c/(a+b)| < 2
-2 < 1 + c/(a+b) < 2
-3 < c/(a+b) < 1
Since a>b>c>0, (a+b) is positive. Therefore, we can multiply by (a+b) without changing the inequality signs.
-3(a+b) < c < a+b
From a>b>c>0, we know that a+b>0 and c>0. Thus, -3(a+b) < c implies that c > -3(a+b), which is always true since c>0 and a+b>0.
The inequality c < a+b implies that a+b-c > 0.
Therefore, a+b-c > 0 is the correct condition.