Prove Triangle Inequality

[nycmathguy]

Prove the triangle inequality showing each step along the way.

Thanks

 

[MathLover1]

AB + BC must be greater than AC, or AB + BC > AC.
AB + AC must be greater than BC, or AB + AC > BC
BC + AC must be greater than AB, or BC + AC > AB.

Triangle Inequality Proof

We need to prove that AB + AC > BC.

Proof: Extend BA to point D such that AD = AC, and join C to D, as shown below:

We note that ∠ACD = ∠D, which means that in ∆ BCD, ∠BCD > ∠D. Sides opposite larger angles are larger, and thus: BD > BC

AB + AD > BC

AB + AC > BC (because AD = AC) This completes our proof.

We can additionally conclude that in a triangle:
Since the sum of any two sides is greater than the third, then the difference of any two sides will be less than the third.
The sum of any two sides must be greater than the third side.
The side opposite to a larger angle is the longest side in the triangle.

 

[nycmathguy]

AB + BC must be greater than AC, or AB + BC > AC.
AB + AC must be greater than BC, or AB + AC > BC
BC + AC must be greater than AB, or BC + AC > AB.

View attachment 463

Triangle Inequality Proof

We need to prove that AB + AC > BC.

Proof: Extend BA to point D such that AD = AC, and join C to D, as shown below:


View attachment 462


We note that ∠ACD = ∠D, which means that in ∆ BCD, ∠BCD > ∠D. Sides opposite larger angles are larger, and thus: BD > BC

AB + AD > BC

AB + AC > BC (because AD = AC) This completes our proof.

We can additionally conclude that in a triangle:
Since the sum of any two sides is greater than the third, then the difference of any two sides will be less than the third.
The sum of any two sides must be greater than the third side.
The side opposite to a larger angle is the longest side in the triangle.

Superbly done. Great work!!
There is another unanswered question in the Algebra forum entitled Short Pythagorean Theorem Proof. The steps are given by David Cohen but slightly hard for me to grasp.