With the on-going trend, Algebraic Inequalities are almost erasing the existence of Geometrical Inequalities . Nevertheless, They are equally beautiful (if not more) , although they seem a lot harder to me. Yes, they can mostly be proved by brute-force trigononmetry or complex (whatever it is , they are equally good solutions) , but I am going to focus on something a bit different to both .
First things first – You can’t multiply, if you can’t add. –
If are the side-lengths of triangle ABC , then the following is true –
Note : The equality case is when they triangle is degenerate or are collinear.
Also, we have the following important and useful observation –
The side opposite the greatest angle is greater.
Mathematically , If then ,
So , If we can find it three positive reals satisfying our triangle inequality, then it is safe to conclude that they are the side-lengths of a triangle.
Surprisingly, quite a few good problems can be solved by this fact alone.
1. If are positive numbers such that – , Then show that –
2. If are positive reals, Prove that –
3. If are side-lengths of a triangle , Prove that –
4. If are side-lengths of a triangle, Prove that –
(Source : IMO 1983)
Hints and Solutions
1. Case work , Trignonmetry
2. Cosine rule
3. Triangle Inequality –
Let be the maximum side of the triangle.
By Triangle Inequality –
Also, (why ?)
4. By Triangle Inequality , $latexb \ge a-c$
Constructing, analogous inequalities and using them in the LHS we get –
The last inequality is Schur’s Inequality of degree 1.