Question

Prove that for any two real numbers x and y, |x + y| ≤ |x| + |y|. Hint: Use the previously proven facts that for any real number a, |a|≥ a and |a|≥−a. You should need only two cases.

Answer 1

AIM: |x + y| ≤ |x| + |y|

x and y are real numbers, then following two cases arises :

CASE 1 : assume x >= 0 and y >= 0

Proof :

As x >= 0 and y >= 0, therefore |x| = x and |y| = y

So LHS(left hand side) = |x| + |y| = x + y

RHS(right hand side) = |x + y| = x + y

Hence proved as x and y are both greater than zero then their modulus can be dropped on right hand side.

This case also implies that for x < 0 and y < 0 the statement holds true.

CASE 2 : assume x >= 0 and y < 0

Proof :

As x >= 0 and y < 0, therefore |x| = x and |y| = -y

and y < 0 means y is negative i.e. y = -y so |-y| = y

Now, two possibilities arise,

1. left side have x >= |y| then

|x + y| = |x| - |y| which is less than or equal to right side |x| + |y|.

2. left side have x < |y| then

|x + y| = |y| - |x| which is less than or equal to right side |x| + |y|.

Hence proved.

This case also implies that when x < 0 and y >= 0 then statement also holds true.