My Math Forum Fundamental and trivial question on triangle inequality.

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 August 15th, 2011, 03:26 AM #1 Newbie   Joined: Jan 2010 Posts: 8 Thanks: 0 Fundamental and trivial question on triangle inequality. It's surprising (for me) that I will ask this but I have never met this. It's well known (e.g Recent Advances in Geometric Inequalities, Mitrinovic et al) that the following is true: [color=#0000BF]A,B,C are sides of a triangle[/color] if and only if [color=#0000BF]A>0, B>0, C>0, A+B>C, A+C>B, C+B>A[/color] Of course the $\Rightarrow$ 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 geometry in the initial chapters, as also being followed with the simple proof I've mentioned. But what about the $\Leftarrow$ part of the equivalence? 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 following theorem as also a reference for the proof: [color=#0000BF]If A>0, B>0, C>0, A+B>C, A+C>B, C+B>A[/color] then [color=#0000BF]a triangle can be constructed with sides A, B, C.[/color] **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. Thanks in advance.
 August 15th, 2011, 04:12 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Fundamental and trivial question on triangle inequality. The triangle with vertices at (0,0), (A,0) and (x,y) where $x=\frac{A^2+C^2-B^2}{2A},$ $y=\frac1{2A}\left((A^2+B^2+C^2)^2-2(A^4+B^4+C^4)\right)^{\frac12}$ has sides with length A, B and C. If you can show that the conditions on side length imply that $y$ is real, then you're done - I'm sure this isn't too tricky, but I'll leave it as an exercise (i.e. I couldn't be bothered right now )
 August 15th, 2011, 04:23 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,753 Thanks: 2136 Using the letters a, b and c for the sides, where a ? b ? c, draw a side of length c, then construct sides with lengths a and b in the usual way. It's easy to show the construction must succeed in producing a triangle if a + b > c.
August 15th, 2011, 04:37 AM   #4
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Re:

Quote:
 Originally Posted by skipjack Using the letters a, b and c for the sides, where a ? b ? c, draw a side of length c, then construct sides with lengths a and b in the usual way. It's easy to show the construction must succeed in producing a triangle if a + b > c.
Not easy for me, in a formal strict way proof of course....

Quote:
 Originally Posted by mattpi The triangle with vertices at (0,0), (A,0) and (x,y) where [color=#FF0000]***********[/color] $x=\frac{A^2+C^2-B^2}{2A},$ [color=#FF0000]***********[/color] $y=\frac1{2A}\left((A^2+B^2+C^2)^2-2(A^4+B^4+C^4)\right)^{\frac12}$ has sides with length A, B and C. If you can show that the conditions on side length imply that $y$ is real, then you're done - I'm sure this isn't too tricky, but I'll leave it as an exercise (i.e. I couldn't be bothered right now )
Before messing around with this, i should ask whether the included in red one should say:
$x=\frac{A^2+C^2-B^2}{2AC},$
That would make more sense, but was that really a typo?

 August 15th, 2011, 05:33 AM #5 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Fundamental and trivial question on triangle inequality. No, it's correct as originally stated. Note that if we multiply A, B and C by a single positive parameter then the values of x and y should also increase by this parameter. Note that $\frac{(\lambda A)^2+(\lambda C)^2-(\lambda B)^2}{2(\lambda A)}=\lambda\frac{A^2+C^2-B^2}{2A},$ so the formula is correct in terms of dimensionality.
August 15th, 2011, 09:46 AM   #6
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Re: Fundamental and trivial question on triangle inequality.

Quote:
 Originally Posted by mattpi No, it's correct as originally stated. Note that if we multiply A, B and C by a single positive parameter then the values of x and y should also increase by this parameter. Note that $\frac{(\lambda A)^2+(\lambda C)^2-(\lambda B)^2}{2(\lambda A)}=\lambda\frac{A^2+C^2-B^2}{2A},$ so the formula is correct in terms of dimensionality.
Yes of course you are right, as also it has to be true that the distance of (x,y) to (A,0) and (x,y) to (0,0) should be B and C respectively so only your initial formulas work with this in mind.

So all in all what we have to prove is the validity of the following implication:
$\left\{ {\begin{array}
{A > 0} \\
{B > 0} \\
{C > 0} \\
\end{array}{\rm{ }}AND{\rm{ }}\begin{array}
{A + B > C} \\
{A + C > B} \\
{B + C > A} \\
\end{array}} \right\} \Rightarrow {({A^2} + {B^2} + {C^2})^2} > 2({A^4} + {B^4} + {C^4})$

I can't really seem to manage solving this but i will try a bit more.

But what is more important to me right now is the lack of any reference i'm noticing, of a book or paper about this kind of fundamental theorem and about a proof of it. Such an elementary and important theorem and not being included in geometry books is very bizarre fact for me.
And even worse, even if i see the solution of the above implication i gave, i would still not be 100% satisfied since i will still miss a geometric proof. Or there isn't any? This can't be the case....

August 15th, 2011, 05:18 PM   #7
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Quote:
 Originally Posted by ChessTal Not easy for me, in a formal strict way proof of course....
Just describe how you would construct the triangle by drawing two arcs, etc. Consider what determines whether the arcs intersect.

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