Prove Triangle Inequality

Set 1.2
Question 67

See attachment.

Work out (a) through (d).

 

Proving Triangle Inequality Theorem.
Sum of any two sides in a triangle is greater than the length of the third side.

recall:

properties of absolute value summary

1.
a. for all real numbers |x| ≥0
b. for all real numbers x≤ | x| and -x≤| x|
c. for all real numbers |x|^2=x^2

2. for all real numbers a and b, we have

a.
|ab| =|a|*|b|
|a/b| =|a|/|b| for

b. |a+b| ≤ |a|+|b| (the triangle inequality)


Complete the following steps to prove the triangle inequality.
(a) Let a and b be real numbers.
Which property in the summary box on page 7 tells us that a<=|a| and b<=|b|

property 1. b

(b) Add the two inequalities in part (a) to obtain

a+b<=|a|+|b|
Therefore we have
|a + b| = a + b ≤ |a| + |b|,
which proves the triangle inequality in this case.

(c) In a similar fashion, add the two inequalities and and deduce that
Applying Lemma 1.b with x = −a we know that −a ≤ | − a|.
But because | − a| = |a|, this implies that −a ≤ |a|.
By the same reasoning, −b ≤ |b|.
Adding these two inequalities we deduce that −a − b ≤ |a| + |b| and therefore
|a + b| = −a − b ≤ |a| + |b|.
This proves the triangle inequality in the second case.
Since these cases exhaust all possibilities, this completes the proof


(d) Why do the results in parts (b) and (c) imply that

|a + b| ≤ |a| + |b|

because |a + b| = a + b
|a + b| = a + b or |a + b| = -(a + b)
since a + b < 0, i n this case |a + b| = −a − b= -(a + b)

 

I could have never done this alone. I am very weak when it comes to math proofs.