# Basic Insights on Geometrical Inequalities.

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. –

Triangle Inequality

If $a,b,c$ are the side-lengths of triangle ABC , then the following is true –

$a+b \ge c$

$b+c \ge a$

$c+a \ge b$

Note : The equality case is when they triangle is degenerate or $A,B,C$ are collinear.

Also, we have the following important and useful observation –

The side opposite the greatest angle is greater.

Mathematically , If $\angle A \ge \angle B$ then , $BC \ge AC$

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.

Problems

1. If $a,b,c$ are positive numbers such that – $a^2+b^2-ab=c^2$ , Then show that –

$(a-b)(b-c) \le 0$

2. If $a,b,c$ are positive reals, Prove that –

$\sqrt{a^2+ab+b^2} \le \sqrt{b^2-bc+c^2} + \sqrt{c^2-ac+a^2}$

3. If $a,b,c$ are side-lengths of a triangle , Prove that –

$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} < 2$

4. If $a,b,c$ are side-lengths of a triangle, Prove that –

$a^2b(a-b) + b^2c(b-c)+c^2a(c-a) \ge 0$

(Source : IMO 1983)

Hints and Solutions

1. Case work , Trignonmetry

3. Triangle Inequality –

Let $a$ be the maximum side of the triangle.

By Triangle Inequality  –$\frac{a}{b+c} \le 1$

Also, $\frac{b}{c+a} + \frac{c}{a+b} \le \frac{b}{c+b} + \frac{c}{c+b} = 1$  (why ?)

Hence proved.

4. By Triangle Inequality , $latexb \ge a-c$

Constructing, analogous inequalities and using them in the LHS we get –

$\sum a^2b(a-b) \ge \sum a(a-b)(a-c) \ge 0$

The last inequality is Schur’s Inequality of degree 1.