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Let $P$ be a point inside $\triangle{ABC}$.

Let $AP=R_a$, $BP=R_b$, $CP=R_c$, $AB=c$, $BC=a$ and $CA=b$. Prove that:

$$\frac{R_a}{2a+b}+\frac{R_b}{2b+c}+\frac{R_c}{2c+a}\geq\frac{1}{\sqrt3}$$

I tried to use C-S, but without success because the equality case is $a=b=c$

and $P$ is a center of a triangle, but by the way, if $P\equiv C$ we obtain: $$\sum\limits_{cyc}\frac{R_a}{2a+b}=\frac{b}{2a+b}+\frac{a}{2b+c}>\frac{b}{2a+b}+\frac{a}{a+3b}>\frac{1}{\sqrt3},$$

where the last inequality is true and strong enough. It says that it's very difficult to use C-S here. Thank you!

Cragfelt
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Michael Rozenberg
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  • Seems heavily related to Weber's problem. Using the framework known for Weber's problem given any triangle $ABC$ you can find $P$ such that it minimizes your quantity. I can't quite figure out how to show that this minimizing $P$ results in a value $\geq (3)^{-1/2}$ though. – Nick R Oct 05 '16 at 23:51
  • I don't have a solution but some thoughts which are too long for a comment. This used to be available here, but upon request of the OP was moved. Please see here: https://1drv.ms/b/s!AjnfDSD0Ps9QcesiXRgs4aiYsPI – Andreas Oct 30 '16 at 21:32
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    The OP has a fundamentally different approach to MSE than me. I see MSE as a forum to discuss, i.e. not only full solutions, but also thoughts may be presented. Here are many people around which then can add to this, and together a solution may arise. The OP instead prefers to keep his own thoughts to himself and accepts only full solutions as posts. Respecting that this is the OP's post, I moved thoughts to an external file. – Andreas Oct 30 '16 at 21:42
  • @Andreas Thank you very much! I think it's best to discuss here: http://www.artofproblemsolving.com/community/c6t243f6_inequalities – Michael Rozenberg Oct 31 '16 at 03:29
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    @MichaelRozenberg I like the company at MSE, thank you. This is a free forum. Nobody is asked to leave here other than those who do not play by the rules. Think about it. – Andreas Oct 31 '16 at 07:41
  • @Andreas I think you one of the best solvers, which there are in this forum. Your proof of this problem (http://math.stackexchange.com/questions/1017110) it's not just beautiful, but it's masterpiece, which says that you great master in inequalities. I did not want to say that you must to leave this forum. In this forum there is very big problem. Here not love discussion. Here works only: ask a questions and give an answers. – Michael Rozenberg Oct 31 '16 at 19:48
  • @Andreas And now the main reason. In the forum about inequalities there is heading "unanswered". If there is at least one answer a problem deleted from "unanswered" and people don't have the opportunity to see this inequality and think about this inequality but I want to see a full answer on my problem. Thank you very much again! – Michael Rozenberg Oct 31 '16 at 19:49
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    @MichaelRozenberg You are right with the "unanswered" argument. If people are heading for "unaswered" questions then it is indeed better that discussions are in the comments. However, a counterargument is that people do not read links in comments. This means, they will aim at following paths which other members have shown to lead nowhere. By the way, the same holds for your questions: they are beautiful and hard pieces. Often you write quite stereotypically "I have tried CS, AM-GM and convexity, without sucess." It would be nice to read what precisely did NOT work, to avoid doing that again. – Andreas Nov 01 '16 at 07:27
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    I'd try to use Bottema's inequality (for two triangles). It often works in such cases. – ivan Jan 20 '17 at 13:48

1 Answers1

2

Here is an idea that I think could work.

$\textbf{Disclaimer: }$This is not a solution, but rather a transformation of the problem into a more 'manageable' form.

We locate the triangle in the complex plane with vertices $A= x_1+\Bbb i x_2,\; B= y_1+\Bbb i y_2$ and $C= z_1+\Bbb i z_2.$ Whitout loss of generalization $P=0.$ We now have

$$a=|B-C|,\; b=|C-A|,\;c=|A-B|,\; R_a= |A|,\; R_b= |B|,\; R_c= |C|.$$ In order to guarantee that $P$ lies inside $\triangle ABC,$ it is necessary and sufficient the existence of scalars $\alpha,\beta,\gamma$ such that:

$$\alpha,\beta,\gamma\geq 0,\;\alpha+\beta+\gamma=1, \; \alpha A+ \beta B+\gamma C=0.$$ The problem is now equivalent to

$$\min f(A,B,C,\alpha,\beta,\gamma)= \frac{|A|}{2|B-C| + |C-A|} + \frac{|B|}{2|C-A| + |A-B|}+ \frac{|C|}{2|A-B| + |B-C|},$$

$$s.t \;\; \alpha,\beta,\gamma\geq 0,\;\alpha+\beta+\gamma=1, \; \alpha A+ \beta B+\gamma C=0.$$ Furthermore, since $f$ is homogeneous, we can assume whitout loss of generality that $|A|^2+|B|^2+|C|^2=1,$(or $|A|+|B|+|C|=1$) so add this constraint to the optimization problem. Now the problem is stated in real terms rather than in geometrical terms, so should make more sense to apply C-S or other known inequalities. Best of lucks!!

John D
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