Question

prove that
|a+b|<=|a|+|b|
@satellite73 @amistre64 @ganeshie8 @myininaya @experimentX @phi @jdoe0001 @thomaster @whpalmer4
just give me idea how to start plz ! anyone !

Answer 1

square it

Answer 2

in other words
\[|a+b|\leq |a|+|b|\] is can be proved showing that
\[(|a+b|)^2\leq (|a|+|b|)^2\]

Answer 3

so i would get finally
(|a+b|)^2 >0 always
=a^2+b^2+2ab

Answer 4

so
\[(a+b)^2=a^2+2ab+b^2=|a|^2+2ab+|b|^2\]
\[\leq |a|^2+2|a||b|+|b|^2=(|a|+|b|)^2\] as desired

Answer 5

there is another simple way
use the inequalities
\[-|a|\leq a\leq |a|\\
-|b|\leq b\leq |b|\] and them up and you get
\[-(|a|+|b|)\leq a+b\leq |a|+|b|\] which is identical so saying
\[|a+b|\leq |a|+|b|\]

Answer 6

ok thats make sense thank you !