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For any two vectors veca and vecb prove ...

For any two vectors `veca and vecb` prove that `|veca+vecb|lt+|veca|+|vecb|`

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The inequality holds trivially in case either `veca=vec0` or `vecb=vec0`. So, let `|veca|ne0ne|vecb|`. Then,
`|vec+vecb|^(2)=(veca+vecb)^(2)=(veca+vecb)^(2).(veca+vecb)`
`=veca.veca+veca.vecb+vecb.veca+vecb.vecb`
`=|veca|^(2)+2veca.vecb+|vecb|^(2)`
`le |veca|^(2)+2|veca.vecb|+|vecb|^(2)" "("Since"x le |x|AA x epsilon "R")`
`le|veca|+2|veca||vecb|+|vecb|^(2)`
`le (|veca|+|vecb|)^(2)`
Hence `" " |veca+vecb| le |veca|+|vecb|`.
Note : If the equality holds in triangle in equality
`|veca+vecb|=|veca|+|vecb|`
then `" " |vecAC|=|vecAB|+|vecBC|`
Showing that the points A,B and C are collinear.
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