Question

|x+y|<|x|+|y| plz solve​

Answer 1

Step-by-step explanation:

To prove this we will use three facts about the absolute value:

1.For all x∈R, |x|²=x²

2.For all x∈R, x≤|x|

3.For all x,y∈R, |xy|=|x||y|

Each of these statements is easily proven from the definition of the absolute value function.

Now, with these three statements the proof is as follows:

Let x,y∈R.

|x−y|²

=(x−y)² (by 1)

=x²−2xy+y²

= |x|²−2xy+|y|² (by 1 again)

≤|x|²−2|xy|+|y|² (by 2)

=|x|²−2|x||y|+|y|² (by 3)

≤|x|²+2|x||y|+|y|² (since by adding 4|x||y| we add a non-negative number)

=(|x|+|y|)²

So, from above we can see that

|x−y|²≤(|x|+|y|)²

So by taking a square root on both sides we get that

||x−y||≤||x|+|y||

⇒|x−y|≤|x|+|y|