In `triangle ABC` internal angle bisector AI,BI and CI are produced to meet opposite sides in `A',B',C'` respectively. Prove that the maximum value of `(AIxxBIxxCI)/(A A'xxBB'xxCC')` is `8/27`

Explanation

Since, angle bisectors divides opposite side in the ratio of sides containing the angle.
`implies BA'=(ac)/(b+c) and CA'=(ab)/(a+c)`
now, BI is also angle bisector of `angleB` for `DeltaABA'`
`implies (AI)/(AI')=(b+c)/(a) implies (AI)/(A A')=(b+c)/(a+b+c)`

Similarly, `(BI)/(B B')=(a+c)/(a+b+c)`
and `(CI)/(C C')=(a+b)/(a+b+c)`
`implies (AI*BI*CI)/(A A'*B B'*C C')=((b+c)(a+b)(a+b))/((a+b+c)(a+b+c)(a+b+c))` . . . (i)
As we know `AB ge GM`, we get
`((b+c)/(a+b+c)+(c+a)/(a+b+c)+(a+b)/(a+b+c))/(3) ge [((a+b)(b+c)(c+a))/((a+b+c)^(3))]^((1)/(3))`
`implies (2(a+b+c))/(3(a+b+c)) ge ([(a+b)(b+c)(c+a)]^(1//3))/(a+b+c)`
`implies ((a+b)(b+c)(c+a))/(a+b+c) le (8)/(27)` . .. (ii)
From eqs. (i) and (ii) we get
`(AI*BI*CI)/(A A'*B B'*C C') le (8)/(27)`.