MathLabs

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⋅BI⋅CI/(AA′⋅BB′⋅CC′)≤8/271/4<AI\cdot BI\cdot CI/(AA'\cdot BB'\cdot CC')\le8/27.
Step 2 of 4: Apply AM-GM for the upper bound
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P=(CA+AB)(AB+BC)(BC+CA)p3≤(23)3=827P=\frac{(CA+AB)(AB+BC)(BC+CA)}{p^3}\le\left(\frac23\right)^3=\frac8{27}
Detailed analysis

The three factors in the numerator sum to 2p, so their arithmetic mean is 2p/3. AM-GM gives their product at most (2p/3)^3. Dividing by p^3 yields the required upper bound.