Date: Aug 22, 2011 8:45 AM
Author: Avni Pllana
Subject: Re: Fundamental and trivial question on triangle inequality.
> It's surprising (for me) that I will ask this but I
> have never met this problem before.
> It's well known (e.g Recent Advances in Geometric
> Inequalities, Mitrinovic et al) that the following
> equivalence is true:
> A,B,C are sides of a triangle if and only if A>0,
> B>0, C>0, A+B>C, A+C>B, C+B>A.
> Of course "=>" the part of the above equivalence is
> well known and it has many proofs and also a simple
> geometric one that Euclid gave ........ all these are
> well known.
> You will find this implication in all books of
> f geometry in the initial chapters, as also being
> followed with the simple proof I've mentioned.
> But what about the "<=" part of the equivalence? I
> I have never seen a proof for this. Can anyone
> provide one, as also a reference for it(a book or
> paper for example)? As crazy as it looks, but looking
> the half internet didn't result in anything! :(
> So to be clear I'm speaking about proving the
> e following theorem as also a reference for the
> If A>0, B>0, C>0, A+B>C, A+C>B, C+B>A then a triangle
> can be constructed with sides A, B, C.
> **By saying "constructed" above, I don't obviously
> mean with compass and ruler construction, but I'm
> referring to the existence of a triangle with sides
> A, B, C.
> For a better viewed version of this(using latex for
> r better viewing of the equations and bold text) as
> also for some suggested solutions see here:
> There, someone gave me an incomplete solution.
> So as it seems, it suffices to show that the
> e following implication is true:
> A>0, B>0, C>0, A+B>C, A+C>B, B+C>A =>
> (A^2 + B^2 + C^2)^2 > 2·(A^4 + B^4 + C^4)
> But i can't really seem to show that also. :(
> But what impresses me most, is the lack of any
> reference i'm noticing(i have posted this in 3 big
> math forums and i got zero replies about any
> references), of a book or paper about this kind of
> fundamental theorem and a proof of it. Such an
> elementary and important theorem and not being
> included in geometry books is very bizarre fact for
> Moreover the lack of a (strict and not descriptive of
> course) geometric proof of it is also odd for me.
> Thanks in advance.
We start from
A>0, B>0, C>0, A+B>C, A+C>B, B+C>A =>
(A^2 + B^2 + C^2)^2 > 2·(A^4 + B^4 + C^4) ...(1)
If we rearrange the last inequality, we obtain
A^4 + B^4 + C^4 < 2*A^2*B^2 + 2*B^2*C^2 + 2*C^2*A^2 ...(2)
From the Law of Cosines we have
A^2 = B^2 + C^2 -2*B*C*cos(A) , or
A^2 - B^2 - C^2 = -2*B*C*cos(A) ... (3)
Squareing both sides of (3), we have
(A^2 - B^2 - C^2)^2 = 4*B^2*C^2*(cos(A))^2 , or
(A^2 - B^2 - C^2)^2 < 4*B^2*C^2 ,...(4)
since (cos(A))^2 < 1.
Finally from (4) follows (2) and (1).