Geometric Maxima and Minimas

The incenter of a triangle divides the angle bisectors into a certain ratio.

Let, $AB = c, BC = a, AC = b$ .

Then, $\frac{AI}{IA'}=\frac{b+c}{a}$ . [Proof is at the end.]

From which follows our desired ratio $\frac{AI}{AA'} = \frac{b+c}{a+b+c}$ .

Similarly, $\frac{BI}{BB'} = \frac{c+a}{a+b+c}$ and $\frac{CI}{CC'} = \frac{a+b}{a+b+c}$ .

So, $\xi := \frac{AI.BI.CI}{AA'.BB'.CC'} = \frac{b+c}{a+b+c} \times \frac{c+a}{a+b+c} \times \frac{a+b}{a+b+c}$

Now, applying AM-GM yields, $\frac{\frac{b+c}{a+b+c} + \frac{c+a}{a+b+c} + \frac{a+b}{a+b+c}}{3} \geq \sqrt[3]{\xi}$ or, $\frac{8}{27} \geq \xi$

Notice, the equality can be achieved when the triangle is equilateral; i.e., $a=b=c$ .

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P.S.

Let, the inradius be $r$ . So, $\triangle AIB = \frac{1}{2} cr$ ; $\triangle BIC = \frac{1}{2} ar$ ; $\triangle CIA = \frac{1}{2} br$ .

Therefore, $\triangle ABC = \frac{1}{2} (a+b+c)r$ . And, $\frac{AA'}{IA'} = \frac{\triangle ABC}{\triangle BIC} = \frac{a+b+c}{a}$ . Some algebraic manipulation will then lead you to $\frac{AI}{IA'}=\frac{b+c}{a}$ .