Question

prove that
|a+b|<=|a|+|b|

Answer 1

@ganeshie8 @myininaya @phi @jdoe0001 @whpalmer4

Answer 2

you don't say what a and b are ?

Answer 3

if they are real numbers, or are vectors, you might do this
| a + b |^2 = (a+b) (a+b) = a^2 + b^2 + 2 a b
if a and b are the same sign (a,b are real numbers) then a b = |a| |b|
if different signs, a b < |a| |b|
if a and b are vectors then \( a \cdot b ≤ |a| |b| \) by Cauchy-Schwartz

in any case
a^2 + b^2 + 2 a b ≤ |a|^2 + |b|^2 + 2 |a| |b| = ( |a| + |b| )^2

or
\[ | a + b |^2 ≤ ( |a| + |b| )^2 \\ |a +b| ≤ |a| + |b| \]