Track / Stage 5 / 373 of 400 #973 of 1964

Problem 973

AIME late

Algebra Difficulty 5.7 Prove it

Given $\triangle ABC$ , with $I$ as its incenter, the internal angle bisectors of $\angle A$ , $\angle B$ , and $\angle C$ intersect their opposite sides at $A'$ , $B'$ , and $C'$ , respectively. Prove:
$\begin{aligned} \frac{5}{4}< & \frac{A I \cdot B I}{A A' \cdot B B'}+\frac{B I \cdot C I}{B B' \cdot C C'} \\ & +\frac{C I \cdot A I}{C C' \cdot A A'}-2 \cdot \frac{I A' \cdot I B' \cdot I C'}{A A' \cdot B B' \cdot C C'} \\ \leqslant & \frac{34}{27} . \end{aligned}$
(Su Xiaoyang, Yuantong Middle School, Yingzhou City, Sichuan, 611236)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Proof: Let $BC = a$ , $CA = b$ , $AB = c$ , and assume $a + b + c = 1$ . Then it is easy to see that
$\begin{array}{l} \frac{AI}{AA'} = b + c, \quad \frac{BI}{BB'} = c + a, \\ \frac{CI}{CC'} = a + b, \quad \frac{IA'}{AA'} = a, \\ \frac{IB'}{BB'} = b, \quad \frac{IC'}{CC'} = c. \end{array}$

Therefore, we only need to prove:
$\begin{array}{c} \frac{5}{4} < (b+c)(c+a) + (c+a)(a+b) \\ + (a+b)(b+c) - 2abc \\ \leq \frac{34}{27} \\ \Leftrightarrow \frac{1}{4} < ab + bc + ca - 2abc \\ \leq \frac{7}{27} \end{array}$

Below, we prove that (1) holds.
In fact, it is easy to know that $0 < a, b, c < \frac{1}{2}$ , so we have
$\begin{array}{l} 0 < (1-2a)(1-2b)(1-2c) \\ \leq \frac{1}{27}[(1-2a) + (1-2b) + (1-2c)]^3. \end{array}$

Expanding and rearranging this inequality, we obtain (1). Therefore, the original inequality holds.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.