Track / Stage 6 / 124 of 400 #1124 of 1964

Problem 1124

National olympiad, first round

Algebra Difficulty 6.2 Prove it

3. From an internal point of a given triangle, three lines are drawn which divide the triangle into 6 parts. Let $S_{1}, S_{2}$ , and $S_{3}$ be the areas of any three of these parts. Prove that:
a) $\frac{1}{S_{1}}+\frac{1}{S_{2}}>\frac{4}{S}$

b) $\frac{1}{S_{1}}+\frac{1}{S_{2}}+\frac{1}{S_{3}}>\frac{9}{S}$ .

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

Solution. b) First method. If we use the fact that $S>S_{1}+S_{2}+S_{3}$ and the inequality between the arithmetic and harmonic mean of the numbers $S_{1}, S_{2}$ , and $S_{3}$ , we get

$\frac{S}{3}>\frac{S_{1}+S_{2}+S_{3}}{3}>\frac{3}{\frac{1}{S_{1}}+\frac{1}{S_{2}}+\frac{1}{S_{3}}}$

from which the desired inequality is obtained.

Second method. For every positive real number $a$ , it holds that $a+\frac{1}{a} \geq 2$ . Now, if we use the fact that $S>S_{1}+S_{2}+S_{3}$ , we get

$\begin{aligned} S\left(\frac{1}{S_{1}}+\frac{1}{S_{2}}+\frac{1}{S_{3}}\right) & >\left(S_{1}+S_{2}+S_{3}\right)\left(\frac{1}{S_{1}}+\frac{1}{S_{2}}+\frac{1}{S_{3}}\right) \\ & =3+\frac{S_{2}}{S_{1}}+\frac{S_{1}}{S_{2}}+\frac{S_{1}}{S_{3}}+\frac{S_{3}}{S_{1}}+\frac{S_{3}}{S_{2}}+\frac{S_{2}}{S_{3}} \leq 9 \end{aligned}$

from which the desired inequality is obtained.

The inequality under a) is proved in a similar manner.

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