Question

Prove that |x+y| <= |x| + |y|

Answer 1

I can Verify!

Answer 2

This is known as triangle inequality,

Here is a proof:

\[|a+b|^2 = (a+b)^2 \] \[= a^2 +2ab+b^2 \le a^2+2|a||b| + b^2\] \[ = a^2 +2ab+b^2 \] \[ = |a|^2+2|a||b| + |b|^2=(|a|+|b|)^2\] \[ \Rightarrow |a+b| <= |a| + |b|\]

Answer 3

For a more deep reference see here : http://en.wikipedia.org/wiki/Triangle_inequality#Normed_vector_space

Answer 4

sqrt((x+y)^2)<=sqrt(x^2)+sqrt(y^2)