Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 7.2 National olympiad, round 2 Prove it

114. The incircle of $\triangle ABC$ touches the sides $BC$ , $CA$ , $AB$ at $A_1$ , $B_1$ , $C_1$ respectively. Let $I_1$ , $I_2$ , $I_3$ be the lengths of the shorter arcs $B_1C_1$ , $C_1A_1$ , $A_1B_1$ respectively. Denote the lengths of the sides $BC$ , $CA$ , $AB$ of $\triangle ABC$ as $a$ , $b$ , $c$ respectively. Prove that: $\frac{a}{I_1} + \frac{b}{I_2} + \frac{c}{I_3} \geq \frac{9 \sqrt{3}}{\pi}$ . (1997 Bosnian Mathematical Olympiad Problem)

Solution

114. Let the inradius of $\triangle ABC$ be $r$ , it is easy to get $I_{1}=r(\pi-A), I_{2}=r(\pi-B), I_{3}=$ $r(\pi-C)$ , thus $\frac{a}{I_{1}}+\frac{b}{I_{2}}+\frac{c}{I_{3}} \geqslant \frac{9 \sqrt{3}}{\pi}$ is equivalent to
$\frac{a}{\pi-A}+\frac{b}{\pi-B}+\frac{c}{\pi-C} \geqslant \frac{9 \sqrt{3}}{\pi} r$

By Chebyshev's inequality, we have
$\frac{a}{\pi-A}+\frac{b}{\pi-B}+\frac{c}{\pi-C} \geqslant \frac{1}{3}(a+b+c)\left(\frac{1}{\pi-A}+\frac{1}{\pi-B}+\frac{1}{\pi-C}\right)$

By Cauchy-Schwarz inequality, we have
$[(\pi-A)+(\pi-B)+(\pi-C)]\left(\frac{1}{\pi-A}+\frac{1}{\pi-B}+\frac{1}{\pi-C}\right) \geqslant 9$

which means
$\frac{1}{\pi-A}+\frac{1}{\pi-B}+\frac{1}{\pi-C} \geqslant \frac{9}{2 \pi}$

Therefore, it suffices to prove $a+b+c \geqslant 6 r$ , which has already been proven in problem 106.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.