Problem 1

Given a triangle ABC, let I be the incenter. The internal bisectors of angles A, B, C meet the opposite sides in A', B', C' respectively. Prove that

1/4<AI\cdot BI\cdot CI/(AA'\cdot BB'\cdot CC')\le8/27

Step 4 of 4: Expand the positive factors for the strict lower bound

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P=\left(\frac12+\frac{x}{p}\right)\left(\frac12+\frac{y}{p}\right)\left(\frac12+\frac{z}{p}\right)>\frac18+\frac{x+y+z}{4p}=\frac14

Detailed analysis

Substitution gives the displayed product. On expanding it, the constant term is 1/8, the linear terms sum to (x+y+z)/(4p), and all remaining terms are positive. Since p=2(x+y+z), the first two contributions already equal 1/4, and positivity is strict. Thus 1/4<P≤8/27.