Proving that the sum of any two sides of the triangle is greater than the third side?

Question

I know it's easy to prove with the help of linear inequalities, but this time I want to prove it with the help of trigonometry. Is it possible? If yes, then how?

Answer 1

Triangle setup: a on the left, C on top, B on the right;

Angle at A is Theta; Angle at B is Lambda

Drops the perpendicular height from C to AB;

M is the intersection of the height from C to AB.

distance x is from A to M and so M to B must be c-x.

THat is, AM=x and MB = c-x

cos(lambda) = (c-x)/a ----> c-x = a*cos(lambda)

cos(theta) = x/b ---> x = b*cos(theta)

Then c = c-x + x = (c-x) + x = a* cos(lambda) + b*cos(theta) <--- substitution

since the side measures of the triangle are guaranteed to be positive, c>0, b>0, and a>0.

So then,

c = |c| = | a* cos(lambda) + b*cos(theta) | <--- takes the absolute value of both sides

<= | a * cos(lambda) | + | b * cos(theta) | <--- Cauchy's Triangle Inequality

= |a| * | cos(lambda)| + |b| * | cos(theta) | <---- property of absolute value : |x*y|=|x|*|y|

Now, the cosine function is STRICTLY between -1 and 1, so the absolute value is AT MOST ONE(1).

So the chain continues as:

<= |a| * 1 + |b|*1 = |a| + |b| = a + b

So, c <= a+b

Now, by contradiction, IF c = a+b, then c - b = a and the pythagorean theorem says:

x^2 + h^2 = b^2

(c-x)^2 + h^2 = a^2

Subtracting them, (c-x)^2 - x^2 = a^2 - b^2

c^2 - 2cx + x^2 - x^2 = a^2 - b^2

c^2 - 2cx = a^2 - b^2

(a+b)^2 - 2cx = a^2 - b^2 <--- substitutes c = a+b

a^2 + 2ab + b^2 - 2cx = a^2 - b^2 <-- FOIL

2ab + b^2 - 2cx = -b^2 <--- a^2 cancels

2ab - 2cx + 2b^2 = 0 <--- moves -b^2 to left side

ab - cx + b^2 = 0 <--- divides by 2

(c-b)*b - cx + b^2 = 0 <--- substitutes a = c-b

bc - b^2 - cx + b^2 = 0

bc - cx = 0

c(b-x) = 0

So then c= 0 or b-x = 0

Since c>0, then b-x = 0 must hold.

Then b = x, which forces cos(theta) = x/b = b/b = 1

The cosine function is one when the angle is 180 degrees.

This is a contradiction since a triangle only has 180 degrees, and

it is supposed to be a right triangle.

Therefore, c < a+b and this concludes the rigorous proof