WLOG let $a \ge b \ge c$. It's equivalent to show that $d^2 \le 9 r^2$ where $d$ is the length of the $A$ bisector. This rewrites as \[ \frac{bc \cdot (b+c-a)(b+c+a)}{(b+c)^2} \le \frac{9}{4} \frac{(b+c-a)(c+a-b)(a+b-c)}{a+b+c} \]Let $a = p+q+r, b = q+r, c = r$ and this simplifies as showing $9(p+r)(p+2q+r)(q+2r)^2 - 4r(q+r)(p+2q+3r)^2 = 9 p^2 q^2 + 18 p q^3 + 32 p^2 q r + 74 p q^2 r + 2 q^3 r + 32 p^2 r^2 + 104 p q r^2 + 17 q^2 r^2 + 48 p r^3 + 24 q r^3 \ge 0$ which is obvious. Sonnhard.