Fundamental and trivial question on triangle inequality.

It's surprising(for me) that i will ask this but i have never met this. (Nerd)

It's well known(e.g Recent Advances in Geometric Inequalities, Mitrinovic, et al) that the following 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 $\displaystyle \Rightarrow$ 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 $\displaystyle \Leftarrow$ 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! (Thinking)(Crying)

So to be clear i'm speaking about proving the following theorem as also a reference for the proof:

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.

Thanks in advance.

Re: Fundamental and trivial question on triangle inequality.

Hi Chess Tal,

The shortest distance between two points is a straight line

Re: Fundamental and trivial question on triangle inequality.

I'm probably missing something but:

say 1 + 1 = 2 is proven ; then no need to prove 2 = 1 + 1

Re: Fundamental and trivial question on triangle inequality.

Quote:

Originally Posted by

**Wilmer** I'm probably missing something but:

say 1 + 1 = 2 is proven ; then no need to prove 2 = 1 + 1

I don't get what you're saying.

I'm not trying to prove 2=1+1.

I have a double implication $\displaystyle A \Leftrightarrow B $ and i know the proof of $\displaystyle A \Rightarrow B $ and i want to prove the $\displaystyle A \Leftarrow B $, i.e(better written) $\displaystyle B \Rightarrow A $.

Quote:

Originally Posted by

**bjhopper** Hi Chess Tal,

The shortest distance between two points is a straight line

So? I don't see how this helps.

Re: Fundamental and trivial question on triangle inequality.

Chess Tal,

DOn't you see that this statement proves that the sum of any two sides of a triangle exceeds the third side?