How can I prove the triangle inequality theorem?

Question

For a secondary school student, what are the most appropriate ways to prove triangle inequality theorem? Please tell me as many ways as you can.

Answer 1

There is an activity you could try. Try it:

Take 9 sticks. Make three different sets such that:

In Set 1, the sum of the length of two sticks is smaller than the length of the third.
In Set 2, the sum of the length of two sticks is equal to the length of the third.
In Set 3, the sum of the length of two stick is larger than the length of the third.

Try making triangles with each of the 3 sets. Only Set 3 gives a triangle.

Answer 2

Prove The Triangle Inequality Theorem

Answer 3

The triangle inequality theorem states that the sum of the measures of any sides of any triangle is greater than the measure of the third side. Simply, you cannot make a triangle if the sum of the two sides is equal or less than the third side.

Example : Sides 2, 3 & 4
2 + 3 = 5 > 4
2 + 4 = 6 > 3
3 + 4 = 7 > 2

You cannot make a triangle with these sides 2, 3 & 5 because 2 + 3 = 5 which is equal to the third side 5

try to draw a triangle with these sides and find out if you can make a triangle

Answer 4

i assume you meant for cosine rule and sine rule

proving cosine rule:
let say a triangle with three points of A, B and C
draw a line, "h" from the top of the triangle of point C and joined it down to the line AB (c*), so the perpendicular line that joined up will be CD with a new point of D. let AC = b, BC = a and AB = c

CD = h
AD = x
BD = AB - x = c - x
Δ CAD = b^2 = h^2 + x^2 ......................(1)
Δ CBD =a^2 = h^2 + (c - x)^2 ........................(2)
(2) - (1) a^2 - b^2 = (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 + c^2 - 2cx ........................(3)
the angle of CAD = cosA = x/b
x = b cosA .........................(4)
(4) replace to (3) a^2 = b^2 + c^2 - 2bc cosA (Proven)

proving sine rule:
take the same triangle as the cosine rule with, AC = b, AB = c, BC = a and CD = h

ΔADC
sinA = h/b
h = b sinA ......................(1)
ΔΒΟC
sin B = h/a
h = a sin B .......................(2)
(1) = (2)
b sin A = a sin B
a/ sin A = b sin B ( Proven)

Answer 5

consider;
-|x|<= x <= |x|
this is true because x either is
-|x| or |x|
so the following is true
-|y|<=y<=|y|
now add them together
-(|x|+|y|)<=x+y<=|x|+|y|
remember the properties
p4) |x|=a if and only if x=(+-)a
p5) |x| this means you can substitute like looking parts of our equation into those properties
substitiute x+y for the x in p4 and p5
substitute |x|+|y| for the a in p4 and p5
now put them together and you have your triangle inequality

|x+y|<=|x|+|y|

I like the geometric proofs but they are difficult to do here because they are done in pictures.