For any 2 vectors a and b , |a| + |b| is always greater than or equal to |a+b| .

Can someone pls explain the statement above

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- Oct 2nd 2008, 08:37 AMmathaddictvector
For any 2 vectors a and b , |a| + |b| is always greater than or equal to |a+b| .

Can someone pls explain the statement above - Oct 2nd 2008, 10:39 AMkbartlett
|a| means the modulus of the vector a, or basically if it is postive it stays the same, and if it is negative u multiply it by -1.

If $\displaystyle a,b \leq 0$ then:

|a|+|b| = |a+b|

if $\displaystyle a,b \geq 0$ then:

|a|+|b| = |a+b|

However if $\displaystyle a < 0$ and $\displaystyle b > 0$ or $\displaystyle b < 0$ and $\displaystyle a > 0$ then:

|a|+|b| > |a+b|

So thats why For any 2 vectors a and b , |a| + |b| is always greater than or equal to |a+b| . - Oct 2nd 2008, 12:23 PMThePerfectHacker
You can prove this by Cauchy-Schwartz inequality: $\displaystyle |\bold{a}\cdot \bold{b}| \leq |\bold{a}|\cdot |\bold{b}|$.

$\displaystyle |\bold{a}+\bold{b}|^2 = (\bold{a}+\bold{b})\cdot (\bold{a}+\bold{b}) = |\bold{a}|^2 + |\bold{b}|^2 + 2\bold{a}\cdot \bold{b} \leq (|\bold{a}|+|\bold{b}|)^2 $

Now take square roots to get,

$\displaystyle |\bold{a}+\bold{b}| \leq |\bold{a}|+|\bold{b}|$