Given a triangle $ABC$, the points $D$, $E$, and $F$ lie on the sides $BC$, $CA$, and $AB$, respectively, are such that $$DC + CE = EA + AF = FB + BD.$$Prove that $$DE + EF + FD \ge \frac12 (AB + BC + CA).$$
Problem
Source: Indonesia INAMO Shortlist 2008 G9
Tags: geometry, Geometric Inequalities, perimeter, semiperimeter
OlympusHero
26.08.2021 05:42
Can I have a hint? I noted that $DC+CE+EA+AF+FB+BD=AB+BC+CA$, so $DC+CE=EA+AF=FB+BD=\frac{AB+BC+CA}{3}$. We also have $DC+CE \geq DE, EA+AF \geq EF, FB+BD \geq FD$, meaning $DE, EF, FD \leq \frac{AB+BC+CA}{3}$, so $DE+EF+FD \leq AB+BC+CA$. However this is not very helpful
OlympusHero
28.08.2021 04:43
Bumping, can I get a hint?
rafaello
11.12.2021 20:23
I found that the minimum of $DE+EF+FD$ occurs when $AD,BE,CF$ are concurrent, though I have no idea how to prove it.
Seicchi28
21.03.2022 19:24
OlympusHero wrote: Bumping, can I get a hint? The idea is similar to this
Let $AB=c, BC=a, CA=b, AE=a_1, AF=a_2, BF=b_1, BD=b_2, CD=c_1, CE=c_2$. Then $a_1 + a_2 = b_1 + b_2 = c_1 + c_2 = \frac{a+b+c}{3}$.
Project $EF$ to $BC$. The length of the projection will be equal to $BC - BF \cos \angle B - CE \cos \angle C = a - b_1 \cos \angle B - c_2 \cos \angle C$. Therefore, $EF \ge a - b_1 \cos \angle B - c_2 \cos \angle C$. Do this similarly to the other two sides to obtain $\sum EF \ge (a+b+c) - \frac{(a+b+c)(\cos \angle A + \cos \angle B + \cos \angle C)}{3}$. We're left to prove $(a+b+c) - \frac{(a+b+c)(\cos \angle A + \cos \angle B + \cos \angle C)}{3} \ge \frac{a+b+c}{2} \iff \frac{3}{2} \ge \sum \cos \angle A$. The last inequality is well known (can be proved e.g. by substituting $\cos \angle A = \frac{a^2 + b^2 - c^2}{2ab}$ and expanding). Equality happens when $a=b=c$, and $EF||BC, DE||AB, DF||AC \implies D, E, F$ are midpoint of $BC, CA, AB$. $\blacksquare$