Condition of existence of a triangle

It's easy to prove that the triangle inequality holds for any triangle with the lengths of sides $a$, $b$ and $c$. But how can one prove that if the triangle inequality holds for any given positives $a$, $b$ and $c$ then a triangle (geometric figure) with the lengths of the sides equal to $a,b$ and $c$ can necessarily be formed?

Solution 1:

Let $B(0,0)$ and $C(a,0)$.

Hence, $BC=a$ and we need to prove that there exists $A(x,y)$ such that $AB=c$ and $AC=b$.

You can write equations of two circles and prove that there are intersection points.

For example $x^2+y^2=c^2$ and $(x-a)^2+y^2=b^2$.

Thus, $-2ax+a^2+c^2=b^2$ or $x=\frac{a^2+c^2-b^2}{2a}$ and $$y^2=c^2-\left(\frac{a^2+c^2-b^2}{2a}\right)^2$$ or $$y^2=\frac{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}{4a^2},$$ which says that there are two intersection points.

Solution 2:

It depends what you are able to assume. Take sides of lengths $a,b,c$ and choose two points $A, B$ distance $c$ apart. Construct a circle of radius $b$ centred at $A$ and a circle of radius $a$ centred at $B$.

The circle centred at $A$ crosses the segment $AB$ extended - at two points. In the direction towards $B$ we have $c-a\lt b\lt c+a$ by the triangle inequality, so the crossing point is within the circle centred at $A$.

For the point away from $B$ we have $c+b\gt a$ so the crossing point is outside the circle centred at $A$.

Therefore the circles must meet (this is where you heed to know what you can assume - in fact at two points). Choose a meeting point and call it $C$. $ABC$ is an example of the triangle you wanted to exist.

Note how all three of the "triangle inequalities" are invoked here.