How would demonstrate |a+b|<= |a|+|b| if a and b are vectors?

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You need to use the following vectors `bar(OA)` and `bar(AB)` that represent the given vectors `bar a` and `bar b` .

Using Chasles's formula in a triangle, yields:

`bar(OA) + bar(AB) = bar(OB)`

Using triangle inequality yields:

`bar(OB) <= |bar(OA) + bar(AB)| <= |bar(OA)| + |bar(AB)|`

Replacing `bar a` for...

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You need to use the following vectors `bar(OA)` and `bar(AB)` that represent the given vectors `bar a` and `bar b` .

Using Chasles's formula in a triangle, yields:

`bar(OA) + bar(AB) = bar(OB)`

Using triangle inequality yields:

`bar(OB) <= |bar(OA) + bar(AB)| <= |bar(OA)| + |bar(AB)|`

Replacing `bar a` for `bar(OA)` and `bar b ` for `bar(AB)` yields:

`|bar a + bar b| <= |bar a| + |bar b|`

Hence, testing if the given vectorial inequality holds, using the Chasles' formula in a triangle, yields that `|bar a + bar b| <= |bar a| + |bar b|` is valid.

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