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 $2\sqrt{2}$. Then

$a > b > c > 0$ $L_1: ax + by + c = 0, \quad L_2: bx + ay + c = 0$ $P = L_1 \cap L_2, \quad \text{distance}\big((1,1), P\big) < 2\sqrt{2}$ $\text{Find: which statement is true}$ $\text{(figure drawn with sample } a=5,\ b=4,\ c=3)$ $1.\ \text{solve } L_1, L_2 \Rightarrow P$ $2.\ \text{distance } AP = ?$ $3.\ AP < 2\sqrt{2} \Rightarrow \text{simplify}$ $L_1 + L_2:\ (a+b)x + (a+b)y + 2c = 0$ $\Rightarrow x + y = -\dfrac{2c}{a+b}$ $L_1 - L_2:\ (a-b)x + (b-a)y = 0$ $\Rightarrow (a-b)(x - y) = 0 \Rightarrow x = y$ $2x = -\dfrac{2c}{a+b} \Rightarrow x = -\dfrac{c}{a+b}$ $P\left(-\dfrac{c}{a+b},\ -\dfrac{c}{a+b}\right)$ $AP = \sqrt{(1-x_P)^2 + (1-y_P)^2}$ $= \sqrt{\left(1+\dfrac{c}{a+b}\right)^2 + \left(1+\dfrac{c}{a+b}\right)^2}$ $AP = \sqrt{2}\left(1 + \dfrac{c}{a+b}\right)$ $= \dfrac{\sqrt{2}\,(a+b+c)}{a+b}$ $AP < 2\sqrt{2}$ $\dfrac{\sqrt{2}\,(a+b+c)}{a+b} < 2\sqrt{2}$ $\dfrac{a+b+c}{a+b} < 2$ $a+b+c < 2a + 2b$ $c < a + b$ $\Rightarrow a + b - c > 0$ $a=5,\ b=4,\ c=3:\quad a+b-c = 6 > 0\ \checkmark$ $AP = \dfrac{\sqrt{2}\cdot 12}{9} \approx 1.886 < 2\sqrt{2}\ \checkmark$ $\boxed{a + b - c > 0} \quad \text{option (1)}$